case study on conic sections

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Conic Sections

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Ayush G Rai
Sarthak Khattar

A conic section is a curve on a plane that is defined by a $2^\text{nd}$-degree polynomial equation in two variables. Conic sections are classified into four groups: parabolas , circles , ellipses , and hyperbolas . Conic sections received their name because they can each be represented by a cross section of a plane cutting through a cone.

\[\begin{array}{ccc}\qquad \quad\text{1: Parabola} \qquad \quad \quad & \text{2: Ellipse and Circle} & \quad \quad \quad \text{3: Hyperbola}\end{array}\]

The practical applications of conic sections are numerous and varied. They are used in physics , orbital mechanics , and optics , among others. In addition to this, each conic section is a locus of points , a set of points that satisfies a condition. Their status as loci of points allows them to be used in practical problems in which the location of an object can vary, but it needs to meet certain conditions. Understanding the coordinate geometry of conic sections allows one to model these situations with the equations of conic sections.

Study Guides > Boundless Algebra

Introduction to conic sections, what are conic sections, learning objectives, key takeaways.

A conic section (or simply conic) is a curve obtained as the intersection of the surface of a cone with a plane; the three types are parabolas, ellipses, and hyperbolas.
A conic section can be graphed on a coordinate plane.
Every conic section has certain features, including at least one focus and directrix. Parabolas have one focus and directrix, while ellipses and hyperbolas have two of each.
A conic section is the set of points [latex]P[/latex] whose distance to the focus is a constant multiple of the distance from [latex]P[/latex] to the directrix of the conic.
vertex : An extreme point on a conic section.
asymptote : A straight line which a curve approaches arbitrarily closely as it goes to infinity.
locus : The set of all points whose coordinates satisfy a given equation or condition.
focus : A point used to construct and define a conic section, at which rays reflected from the curve converge (plural: foci).
nappe : One half of a double cone.
conic section : Any curve formed by the intersection of a plane with a cone of two nappes.
directrix : A line used to construct and define a conic section; a parabola has one directrix; ellipses and hyperbolas have two (plural: directrices).

Defining Conic Sections

An hourglass-shaped double cone (two nappes). A horizontal plane through the cone makes a circle. A plane at an angle less than the edge of the cone makes an ellipse. A plane at the angle of the edge of the cone makes a parabola. A plane at an angle greater than the edge of the cone makes a hyperbola.

Common Parts of Conic Sections

An ellipse is an oblong closed shape and has one focus inside the ellipse to one side along its long axis and its directrix outside it perpendicular to the long axis of the ellipse. A parabola has a focus inside the parabola along its axis of symmetry and a directrix outside it perpendicular to the axis of symmetry. A hyperbola is comprised of two symmetrical open curves with similar shapes to parabolas, but whose branches go towards lines instead of increasing in slope forever. It has a focus inside each curve along the axis of symmetry and a directrix outside each curve perpendicular to the axis of symmetry.

Applications of Conic Sections

Eccentricity.

Eccentricity is a parameter associated with every conic section, and can be thought of as a measure of how much the conic section deviates from being circular.
The eccentricity of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix.
The value of [latex]e[/latex] can be used to determine the type of conic section. If [latex]e= 1[/latex] it is a parabola, if [latex]e < 1[/latex] it is an ellipse, and if [latex]e > 1[/latex] it is a hyperbola.
eccentricity : A parameter of a conic section that describes how much the conic section deviates from being circular.

Defining Eccentricity

If [latex]e = 1[/latex], the conic is a parabola
If [latex]e < 1[/latex], it is an ellipse
If [latex]e > 1[/latex], it is a hyperbola

The ellipse, an oblong closed shape, has a focus inside it along its longest axis with the distances to the edges of the shape radiating out from it. Each of these points has a line going from it to the directrix, a vertical line perpendicular to the longest axis of the ellipse outside the ellipse. All the lines to the directrix are parallel to each other and perpendicular to the directrix. The parabola has a focus inside it along its axis of symmetry, and a directrix outside it perpendicular to the axis of symmetry. Lines radiate out to the parabola from it, and from those points to the directrix perpendicularly. The hyperbola is another open curve that goes out towards linear asymptotes. It has an axis of symmetry down the middle like a parabola, with a focus inside the hyperbola. The directrix is outside the hyperbola perpendicular to the axis of symmetry. Lines radiate from the focus to the hyperbola, and from those points to the directrix perpendicularly.

Conceptualizing Eccentricity

Types of conic sections.

Conic sections are a particular type of shape formed by the intersection of a plane and a right circular cone. Depending on the angle between the plane and the cone, four different intersection shapes can be formed.
The types of conic sections are circles, ellipses, hyperbolas, and parabolas.
Each conic section also has a degenerate form; these take the form of points and lines.
degenerate : A conic section which does not fit the standard form of equation.
asymptote : A line which a curved function or shape approaches but never touches.
hyperbola : The conic section formed by the plane being perpendicular to the base of the cone.
focus : A point away from a curved line, around which the curve bends.
circle : The conic section formed by the plane being parallel to the base of the cone.
ellipse : The conic section formed by the plane being at an angle to the base of the cone.
eccentricity : A dimensionless parameter characterizing the shape of a conic section.
Parabola : The conic section formed by the plane being parallel to the cone.
vertex : The turning point of a curved shape.

Three double cones (two cones touching each other at their points, like an hourglass) with planes crossing through them. 1 has the plane perpendicuar to the edge of one cone, making a parabola. 2 has two planes crossing through it at angles less than the edge of the cone, making one closed ellipse and one circle when the plane is horizontal. 3 has a plane crossing through the cone vertically, making a hyperbola

A vertex, which is the point at which the curve turns around
A focus, which is a point not on the curve about which the curve bends
An axis of symmetry, which is a line connecting the vertex and the focus which divides the parabola into two equal halves
A center point
A radius, which the distance from any point on the circle to the center point

A major axis, which is the longest width across the ellipse
A minor axis, which is the shortest width across the ellipse
A center, which is the intersection of the two axes
Two focal points —for any point on the ellipse, the sum of the distances to both focal points is a constant
Asymptote lines—these are two linear graphs that the curve of the hyperbola approaches, but never touches
A center, which is the intersection of the asymptotes
Two focal points, around which each of the two branches bend
Two vertices, one for each branch

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7.5 Conic Sections

Learning objectives.

7.5.1 Identify the equation of a parabola in standard form with given focus and directrix.
7.5.2 Identify the equation of an ellipse in standard form with given foci.
7.5.3 Identify the equation of a hyperbola in standard form with given foci.
7.5.4 Recognize a parabola, ellipse, or hyperbola from its eccentricity value.
7.5.5 Write the polar equation of a conic section with eccentricity e e .
7.5.6 Identify when a general equation of degree two is a parabola, ellipse, or hyperbola.

Conic sections have been studied since the time of the ancient Greeks, and were considered to be an important mathematical concept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since then, important applications of conic sections have arisen (for example, in astronomy), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even architecture. In this section we discuss the three basic conic sections, some of their properties, and their equations.

Conic sections get their name because they can be generated by intersecting a plane with a cone. A cone has two identically shaped parts called nappes . One nappe is what most people mean by “cone,” having the shape of a party hat. A right circular cone can be generated by revolving a line passing through the origin around the y -axis as shown.

Conic sections are generated by the intersection of a plane with a cone ( Figure 7.44 ). If the plane intersects both nappes, then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than 90 ° ) , 90 ° ) , then the conic section is an ellipse.

A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.

A parabola is the set of all points whose distance from a fixed point, called the focus , is equal to the distance from a fixed line, called the directrix . The point halfway between the focus and the directrix is called the vertex of the parabola.

A graph of a typical parabola appears in Figure 7.45 . Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point P with coordinates ( x 1 , y 1 ) ( x 1 , y 1 ) and point Q with coordinates ( x 2 , y 2 ) , ( x 2 , y 2 ) , the distance between them is given by the formula

Then from the definition of a parabola and Figure 7.45 , we get

Squaring both sides and simplifying yields

Now suppose we want to relocate the vertex. We use the variables ( h , k ) ( h , k ) to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates ( h , k + p ) ( h , k + p ) and the directrix has the equation y = k − p . y = k − p . Going through the same derivation yields the formula ( x − h ) 2 = 4 p ( y − k ) . ( x − h ) 2 = 4 p ( y − k ) . Solving this equation for y leads to the following theorem.

Theorem 7.8

Equations for parabolas.

Given a parabola opening upward with vertex located at ( h , k ) ( h , k ) and focus located at ( h , k + p ) , ( h , k + p ) , where p is a constant, the equation for the parabola is given by

This is the standard form of a parabola.

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

In addition, the equation of a parabola can be written in the general form , though in this form the values of h , k , and p are not immediately recognizable. The general form of a parabola is written as

The first equation represents a parabola that opens either up or down. The second equation represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.

Example 7.19

Converting the equation of a parabola from general into standard form.

Put the equation x 2 − 4 x − 8 y + 12 = 0 x 2 − 4 x − 8 y + 12 = 0 into standard form and graph the resulting parabola.

Since y is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for y, which will put the equation into standard form. To do that, first add 8 y 8 y to both sides of the equation:

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of x and square it. This gives ( −4 2 ) 2 = 4 . ( −4 2 ) 2 = 4 . Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

Now combine like terms and factor the quantity inside the parentheses:

Finally, divide by 8:

This equation is now in standard form. Comparing this to Equation 7.11 gives h = 2 , h = 2 , k = 1 , k = 1 , and p = 2 . p = 2 . The parabola opens up, with vertex at ( 2 , 1 ) , ( 2 , 1 ) , focus at ( 2 , 3 ) , ( 2 , 3 ) , and directrix y = −1 . y = −1 . The graph of this parabola appears as follows.

Checkpoint 7.18

Put the equation 2 y 2 − x + 12 y + 16 = 0 2 y 2 − x + 12 y + 16 = 0 into standard form and graph the resulting parabola.

The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

Consider a parabolic dish designed to collect signals from a satellite in space. The dish is aimed directly at the satellite, and a receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of the parabola to the receiver, which collects and decodes the digital signals. This allows a small receiver to gather signals from a wide angle of sky. Flashlights and headlights in a car work on the same principle, but in reverse: the source of the light (that is, the light bulb) is located at the focus and the reflecting surface on the parabolic mirror focuses the beam straight ahead. This allows a small light bulb to illuminate a wide angle of space in front of the flashlight or car.

An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.

An ellipse is the set of all points for which the sum of their distances from two fixed points (the foci) is constant.

A graph of a typical ellipse is shown in Figure 7.48 . In this figure the foci are labeled as F F and F ′ . F ′ . Both are the same fixed distance from the origin, and this distance is represented by the variable c . Therefore the coordinates of F F are ( c , 0 ) ( c , 0 ) and the coordinates of F ′ F ′ are ( − c , 0 ) . ( − c , 0 ) . The points P P and P ′ P ′ are located at the ends of the major axis of the ellipse, and have coordinates ( a , 0 ) ( a , 0 ) and ( − a , 0 ) , ( − a , 0 ) , respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2 a. Furthermore, P P and P ′ P ′ are called the vertices of the ellipse. The points Q Q and Q ′ Q ′ are located at the ends of the minor axis of the ellipse, and have coordinates ( 0 , b ) ( 0 , b ) and ( 0 , − b ) , ( 0 , − b ) , respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.

According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point P. Since the coordinates of point P are ( a , 0 ) , ( a , 0 ) , the sum of the distances is

Therefore the sum of the distances from an arbitrary point A with coordinates ( x , y ) ( x , y ) is also equal to 2 a. Using the distance formula, we get

Subtract the second radical from both sides and square both sides:

Now isolate the radical on the right-hand side and square again:

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

Divide both sides by a 2 − c 2 . a 2 − c 2 . This gives the equation

If we refer back to Figure 7.48 , then the length of each of the two green line segments is equal to a . This is true because the sum of the distances from the point Q to the foci F and F ′ F and F ′ is equal to 2 a , and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length a and leg lengths b and c . From the Pythagorean theorem, a 2 = b 2 = c 2 a 2 = b 2 = c 2 and b 2 + a 2 − c 2 . b 2 + a 2 − c 2 . Therefore the equation of the ellipse becomes

Finally, if the center of the ellipse is moved from the origin to a point ( h , k ) , ( h , k ) , we have the following standard form of an ellipse.

Theorem 7.9

Equation of an ellipse in standard form.

Consider the ellipse with center ( h , k ) , ( h , k ) , a horizontal major axis with length 2 a , and a vertical minor axis with length 2 b . Then the equation of this ellipse in standard form is

and the foci are located at ( h ± c , k ) , ( h ± c , k ) , where c 2 = a 2 − b 2 . c 2 = a 2 − b 2 . The equations of the directrices are x = h ± a 2 c . x = h ± a 2 c .

If the major axis is vertical, then the equation of the ellipse becomes

and the foci are located at ( h , k ± c ) , ( h , k ± c ) , where c 2 = a 2 − b 2 . c 2 = a 2 − b 2 . The equations of the directrices in this case are y = k ± a 2 c . y = k ± a 2 c .

If the major axis is horizontal, then the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is called vertical. The equation of an ellipse is in general form if it is in the form A x 2 + B y 2 + C x + D y + E = 0 , A x 2 + B y 2 + C x + D y + E = 0 , where A and B are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square.

Example 7.20

Finding the standard form of an ellipse.

Put the equation 9 x 2 + 4 y 2 − 36 x + 24 y + 36 = 0 9 x 2 + 4 y 2 − 36 x + 24 y + 36 = 0 into standard form and graph the resulting ellipse.

First subtract 36 from both sides of the equation:

Next group the x terms together and the y terms together, and factor out the common factor:

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives ( −4 2 ) 2 = 4 . ( −4 2 ) 2 = 4 . In the second set of parentheses, take half the coefficient of y and square it. This gives ( 6 2 ) 2 = 9 . ( 6 2 ) 2 = 9 . Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

Now factor both sets of parentheses and divide by 36:

The equation is now in standard form. Comparing this to Equation 7.14 gives h = 2 , h = 2 , k = −3 , k = −3 , a = 3 , a = 3 , and b = 2 . b = 2 . This is a vertical ellipse with center at ( 2 , −3 ) , ( 2 , −3 ) , major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

Checkpoint 7.19

Put the equation 9 x 2 + 16 y 2 + 18 x − 64 y − 71 = 0 9 x 2 + 16 y 2 + 18 x − 64 y − 71 = 0 into standard form and graph the resulting ellipse.

According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in Figure 7.50 (a). Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth.

Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focus after mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure 7.50 (b). This hall served as the meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they can hear someone standing close by. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Central Station in New York City.

A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.

A hyperbola is the set of all points where the difference between their distances from two fixed points (the foci) is constant.

A graph of a typical hyperbola appears as follows.

The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let P be a point on the hyperbola with coordinates ( x , y ) . ( x , y ) . Then the definition of the hyperbola gives | d ( P , F 1 ) − d ( P , F 2 ) | = constant . | d ( P , F 1 ) − d ( P , F 2 ) | = constant . To simplify the derivation, assume that P is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates ( a , 0 ) , ( a , 0 ) , so

This equation is therefore true for any point on the hyperbola. Returning to the coordinates ( x , y ) ( x , y ) for P :

Add the second radical from both sides and square both sides:

Finally, divide both sides by a 2 − c 2 . a 2 − c 2 . This gives the equation

We now define b so that b 2 = c 2 − a 2 . b 2 = c 2 − a 2 . This is possible because c > a . c > a . Therefore the equation of the ellipse becomes

Finally, if the center of the hyperbola is moved from the origin to the point ( h , k ) , ( h , k ) , we have the following standard form of a hyperbola.

Theorem 7.10

Equation of a hyperbola in standard form.

Consider the hyperbola with center ( h , k ) , ( h , k ) , a horizontal major axis, and a vertical minor axis. Then the equation of this ellipse is

and the foci are located at ( h ± c , k ) , ( h ± c , k ) , where c 2 = a 2 + b 2 . c 2 = a 2 + b 2 . The equations of the asymptotes are given by y = k ± b a ( x − h ) . y = k ± b a ( x − h ) . The equations of the directrices are

If the major axis is vertical, then the equation of the hyperbola becomes

and the foci are located at ( h , k ± c ) , ( h , k ± c ) , where c 2 = a 2 + b 2 . c 2 = a 2 + b 2 . The equations of the asymptotes are given by y = k ± a b ( x − h ) . y = k ± a b ( x − h ) . The equations of the directrices are

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form A x 2 + B y 2 + C x + D y + E = 0 , A x 2 + B y 2 + C x + D y + E = 0 , where A and B have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

Example 7.21

Finding the standard form of a hyperbola.

Put the equation 9 x 2 − 16 y 2 + 36 x + 32 y − 124 = 0 9 x 2 − 16 y 2 + 36 x + 32 y − 124 = 0 into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

First add 124 to both sides of the equation:

Next group the x terms together and the y terms together, then factor out the common factors:

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives ( 4 2 ) 2 = 4 . ( 4 2 ) 2 = 4 . In the second set of parentheses, take half the coefficient of y and square it. This gives ( −2 2 ) 2 = 1 . ( −2 2 ) 2 = 1 . Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

Next factor both sets of parentheses and divide by 144:

The equation is now in standard form. Comparing this to Equation 7.15 gives h = −2 , h = −2 , k = 1 , k = 1 , a = 4 , a = 4 , and b = 3 . b = 3 . This is a horizontal hyperbola with center at ( −2 , 1 ) ( −2 , 1 ) and asymptotes given by the equations y = 1 ± 3 4 ( x + 2 ) . y = 1 ± 3 4 ( x + 2 ) . The graph of this hyperbola appears in the following figure.

Checkpoint 7.20

Put the equation 4 y 2 − 9 x 2 + 16 y + 18 x − 29 = 0 4 y 2 − 9 x 2 + 16 y + 18 x − 29 = 0 into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Hyperbolas also have interesting reflective properties. A ray directed toward one focus of a hyperbola is reflected by a hyperbolic mirror toward the other focus. This concept is illustrated in the following figure.

This property of the hyperbola has important applications. It is used in radio direction finding (since the difference in signals from two towers is constant along hyperbolas), and in the construction of mirrors inside telescopes (to reflect light coming from the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solar system, if the speed is great enough to escape the Sun’s gravitational pull, then the path that the comet takes as it passes through the solar system is hyperbolic.

Eccentricity and Directrix

An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.

The eccentricity e of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix. This value is constant for any conic section, and can define the conic section as well:

If e = 1 , e = 1 , the conic is a parabola.
If e < 1 , e < 1 , it is an ellipse.
If e > 1 , e > 1 , it is a hyperbola.

The eccentricity of a circle is zero. The directrix of a conic section is the line that, together with the point known as the focus, serves to define a conic section. Hyperbolas and noncircular ellipses have two foci and two associated directrices. Parabolas have one focus and one directrix.

The three conic sections with their directrices appear in the following figure.

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are x = ± a 2 c . x = ± a 2 c . The right vertex of the ellipse is located at ( a , 0 ) ( a , 0 ) and the right focus is ( c , 0 ) . ( c , 0 ) . Therefore the distance from the vertex to the focus is a − c a − c and the distance from the vertex to the right directrix is a 2 c − a . a 2 c − a . This gives the eccentricity as

Since c < a , c < a , this step proves that the eccentricity of an ellipse is less than 1. The directrices of a horizontal hyperbola are also located at x = ± a 2 c , x = ± a 2 c , and a similar calculation shows that the eccentricity of a hyperbola is also e = c a . e = c a . However in this case we have c > a , c > a , so the eccentricity of a hyperbola is greater than 1.

Example 7.22

Determining eccentricity of a conic section.

Determine the eccentricity of the ellipse described by the equation

From the equation we see that a = 5 a = 5 and b = 4 . b = 4 . The value of c can be calculated using the equation a 2 = b 2 + c 2 a 2 = b 2 + c 2 for an ellipse. Substituting the values of a and b and solving for c gives c = 3 . c = 3 . Therefore the eccentricity of the ellipse is e = c a = 3 5 = 0.6 . e = c a = 3 5 = 0.6 .

Checkpoint 7.21

Determine the eccentricity of the hyperbola described by the equation

Polar Equations of Conic Sections

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section p is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where a is the length of the semi-major axis (i.e., half the length of the major axis), c is the distance from the origin to the focus, and e is the eccentricity. In the case of a parabola, a represents the distance from the vertex to the focus.

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.

Theorem 7.11

Polar equation of conic sections.

The polar equation of a conic section with focal parameter p is given by

In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., e = 0 ) . e = 0 ) .

Example 7.23

Graphing a conic section in polar coordinates.

Identify and create a graph of the conic section described by the equation

The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter p can be calculated by using the equation e p = 3 . e p = 3 . Since e = 2 , e = 2 , this gives p = 3 2 . p = 3 2 . The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for θ θ and create a table of values. Then we can graph the hyperbola ( Figure 7.55 ).

Checkpoint 7.22

General equations of degree two.

A general equation of degree two can be written in the form

The graph of an equation of this form is a conic section. If B ≠ 0 B ≠ 0 then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section 4 A C − B 2 . 4 A C − B 2 . One of the following cases must be true:

4 A C − B 2 > 0 . 4 A C − B 2 > 0 . If so, the graph is an ellipse.
4 A C − B 2 = 0 . 4 A C − B 2 = 0 . If so, the graph is a parabola.
4 A C − B 2 < 0 . 4 A C − B 2 < 0 . If so, the graph is a hyperbola.

The simplest example of a second-degree equation involving a cross term is x y = 1 . x y = 1 . This equation can be solved for y to obtain y = 1 x . y = 1 x . The graph of this function is called a rectangular hyperbola as shown.

The asymptotes of this hyperbola are the x and y coordinate axes. To determine the angle θ θ of rotation of the conic section, we use the formula cot 2 θ = A − C B . cot 2 θ = A − C B . In this case A = C = 0 A = C = 0 and B = 1 , B = 1 , so cot 2 θ = ( 0 − 0 ) / 1 = 0 cot 2 θ = ( 0 − 0 ) / 1 = 0 and θ = 45 ° . θ = 45 ° . The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled A ′ , B ′ , C ′ , D ′ , E ′ , and F ′ , A ′ , B ′ , C ′ , D ′ , E ′ , and F ′ , and are given by the formulas

The procedure for graphing a rotated conic is the following:

Identify the conic section using the discriminant 4 A C − B 2 . 4 A C − B 2 .
Determine θ θ using the formula cot 2 θ = A − C B . cot 2 θ = A − C B .
Calculate A ′ , B ′ , C ′ , D ′ , E ′ , and F ′ . A ′ , B ′ , C ′ , D ′ , E ′ , and F ′ .
Rewrite the original equation using A ′ , B ′ , C ′ , D ′ , E ′ , and F ′ . A ′ , B ′ , C ′ , D ′ , E ′ , and F ′ .
Draw a graph using the rotated equation.

Example 7.24

Identifying a rotated conic.

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

In this equation, A = 13 , B = −6 3 , C = 7 , D = 0 , E = 0 , A = 13 , B = −6 3 , C = 7 , D = 0 , E = 0 , and F = −256 . F = −256 . The discriminant of this equation is 4 A C − B 2 = 4 ( 13 ) ( 7 ) − ( −6 3 ) 2 = 364 − 108 = 256 . 4 A C − B 2 = 4 ( 13 ) ( 7 ) − ( −6 3 ) 2 = 364 − 108 = 256 . Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use cot 2 θ = A − C B . cot 2 θ = A − C B . This gives

Therefore 2 θ = 120 o 2 θ = 120 o and θ = 60 o , θ = 60 o , which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

The equation of the conic in the rotated coordinate system becomes

A graph of this conic section appears as follows.

Checkpoint 7.23

Section 7.5 exercises.

For the following exercises, determine the equation of the parabola using the information given.

Focus ( 4 , 0 ) ( 4 , 0 ) and directrix x = −4 x = −4

Focus ( 0 , −3 ) ( 0 , −3 ) and directrix y = 3 y = 3

Focus ( 0 , 0.5 ) ( 0 , 0.5 ) and directrix y = −0.5 y = −0.5

Focus ( 2 , 3 ) ( 2 , 3 ) and directrix x = −2 x = −2

Focus ( 0 , 2 ) ( 0 , 2 ) and directrix y = 4 y = 4

Focus ( −1 , 4 ) ( −1 , 4 ) and directrix x = 5 x = 5

Focus ( −3 , 5 ) ( −3 , 5 ) and directrix y = 1 y = 1

Focus ( 5 2 , −4 ) ( 5 2 , −4 ) and directrix x = 7 2 x = 7 2

For the following exercises, determine the equation of the ellipse using the information given.

Endpoints of major axis at ( 4 , 0 ) , ( −4 , 0 ) ( 4 , 0 ) , ( −4 , 0 ) and foci located at ( 2 , 0 ) , ( −2 , 0 ) ( 2 , 0 ) , ( −2 , 0 )

Endpoints of major axis at ( 0 , 5 ) , ( 0 , −5 ) ( 0 , 5 ) , ( 0 , −5 ) and foci located at ( 0 , 3 ) , ( 0 , −3 ) ( 0 , 3 ) , ( 0 , −3 )

Endpoints of minor axis at ( 0 , 2 ) , ( 0 , −2 ) ( 0 , 2 ) , ( 0 , −2 ) and foci located at ( 3 , 0 ) , ( −3 , 0 ) ( 3 , 0 ) , ( −3 , 0 )

Endpoints of major axis at ( −3 , 3 ) , ( 7 , 3 ) ( −3 , 3 ) , ( 7 , 3 ) and foci located at ( −2 , 3 ) , ( 6 , 3 ) ( −2 , 3 ) , ( 6 , 3 )

Endpoints of major axis at ( −3 , 5 ) , ( −3 , −3 ) ( −3 , 5 ) , ( −3 , −3 ) and foci located at ( −3 , 3 ) , ( −3 , −1 ) ( −3 , 3 ) , ( −3 , −1 )

Endpoints of major axis at ( 0 , 0 ) , ( 0 , 4 ) ( 0 , 0 ) , ( 0 , 4 ) and foci located at ( 5 , 2 ) , ( −5 , 2 ) ( 5 , 2 ) , ( −5 , 2 )

Foci located at ( 2 , 0 ) , ( −2 , 0 ) ( 2 , 0 ) , ( −2 , 0 ) and eccentricity of 1 2 1 2

Foci located at ( 0 , −3 ) , ( 0 , 3 ) ( 0 , −3 ) , ( 0 , 3 ) and eccentricity of 3 4 3 4

For the following exercises, determine the equation of the hyperbola using the information given.

Vertices located at ( 5 , 0 ) , ( −5 , 0 ) ( 5 , 0 ) , ( −5 , 0 ) and foci located at ( 6 , 0 ) , ( −6 , 0 ) ( 6 , 0 ) , ( −6 , 0 )

Vertices located at ( 0 , 2 ) , ( 0 , −2 ) ( 0 , 2 ) , ( 0 , −2 ) and foci located at ( 0 , 3 ) , ( 0 , −3 ) ( 0 , 3 ) , ( 0 , −3 )

Endpoints of the conjugate axis located at ( 0 , 3 ) , ( 0 , −3 ) ( 0 , 3 ) , ( 0 , −3 ) and foci located ( 4 , 0 ) , ( −4 , 0 ) ( 4 , 0 ) , ( −4 , 0 )

Vertices located at ( 0 , 1 ) , ( 6 , 1 ) ( 0 , 1 ) , ( 6 , 1 ) and focus located at ( 8 , 1 ) ( 8 , 1 )

Vertices located at ( −2 , 0 ) , ( −2 , −4 ) ( −2 , 0 ) , ( −2 , −4 ) and focus located at ( −2 , −8 ) ( −2 , −8 )

Endpoints of the conjugate axis located at ( 3 , 2 ) , ( 3 , 4 ) ( 3 , 2 ) , ( 3 , 4 ) and focus located at ( 3 , 7 ) ( 3 , 7 )

Foci located at ( - 6 , 0 ) , ( 6 , 0 ) ( - 6 , 0 ) , ( 6 , 0 ) and eccentricity of 3

( 0 , 10 ) , ( 0 , −10 ) ( 0 , 10 ) , ( 0 , −10 ) and eccentricity of 2.5

For the following exercises, consider the following polar equations of conics. Determine the eccentricity and identify the conic.

r = −1 1 + cos θ r = −1 1 + cos θ

r = 8 2 − sin θ r = 8 2 − sin θ

r = 5 2 + sin θ r = 5 2 + sin θ

r = 5 −1 + 2 sin θ r = 5 −1 + 2 sin θ

r = 3 2 − 6 sin θ r = 3 2 − 6 sin θ

r = 3 −4 + 3 sin θ r = 3 −4 + 3 sin θ

For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.

Directrix: x = 4 ; e = 1 5 Directrix: x = 4 ; e = 1 5

Directrix: x = −4 ; e = 5 Directrix: x = −4 ; e = 5

Directrix: y = 2 ; e = 2 Directrix: y = 2 ; e = 2

Directrix: y = −2 ; e = 1 2 Directrix: y = −2 ; e = 1 2

For the following exercises, sketch the graph of each conic.

r = 1 1 + sin θ r = 1 1 + sin θ

r = 1 1 − cos θ r = 1 1 − cos θ

r = 4 1 + cos θ r = 4 1 + cos θ

r = 10 5 + 4 sin θ r = 10 5 + 4 sin θ

r = 15 3 − 2 cos θ r = 15 3 − 2 cos θ

r = 32 3 + 5 sin θ r = 32 3 + 5 sin θ

r ( 2 + sin θ ) = 4 r ( 2 + sin θ ) = 4

r = 3 2 + 6 sin θ r = 3 2 + 6 sin θ

r = 3 −4 + 2 sin θ r = 3 −4 + 2 sin θ

x 2 9 + y 2 4 = 1 x 2 9 + y 2 4 = 1

x 2 4 + y 2 16 = 1 x 2 4 + y 2 16 = 1

4 x 2 + 9 y 2 = 36 4 x 2 + 9 y 2 = 36

25 x 2 − 4 y 2 = 100 25 x 2 − 4 y 2 = 100

x 2 16 − y 2 9 = 1 x 2 16 − y 2 9 = 1

x 2 = 12 y x 2 = 12 y

y 2 = 20 x y 2 = 20 x

12 x = 5 y 2 12 x = 5 y 2

For the following equations, determine which of the conic sections is described.

x y = 4 x y = 4

x 2 + 4 x y − 2 y 2 − 6 = 0 x 2 + 4 x y − 2 y 2 − 6 = 0

x 2 + 2 3 x y + 3 y 2 − 6 = 0 x 2 + 2 3 x y + 3 y 2 − 6 = 0

x 2 − x y + y 2 − 2 = 0 x 2 − x y + y 2 − 2 = 0

34 x 2 − 24 x y + 41 y 2 − 25 = 0 34 x 2 − 24 x y + 41 y 2 − 25 = 0

52 x 2 − 72 x y + 73 y 2 + 40 x + 30 y − 75 = 0 52 x 2 − 72 x y + 73 y 2 + 40 x + 30 y − 75 = 0

The mirror in an automobile headlight has a parabolic cross section, with the lightbulb at the focus. On a schematic, the equation of the parabola is given as x 2 = 4 y . x 2 = 4 y . At what coordinates should you place the lightbulb?

A satellite dish is shaped like a paraboloid of revolution. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?

Consider the satellite dish of the preceding problem. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?

A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.

Whispering galleries are rooms designed with elliptical ceilings. A person standing at one focus can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet and the foci are located 30 feet from the center, find the height of the ceiling at the center.

A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus and the other focus is 80 feet away, what is the length and the height at the center of the gallery?

For the following exercises, determine the polar equation form of the orbit given the length of the major axis and eccentricity for the orbits of the comets or planets. Distance is given in astronomical units (AU).

Halley’s Comet: length of major axis = 35.88, eccentricity = 0.967

Hale-Bopp Comet: length of major axis = 525.91, eccentricity = 0.995

Mars: length of major axis = 3.049, eccentricity = 0.0934

Jupiter: length of major axis = 10.408, eccentricity = 0.0484

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Module 7: Parametric Equations and Polar Coordinates

Introduction to conic sections.

The line y = 3x is drawn and then rotated around the y-axis to create two nappes, that is, a cone that is both above and below the x axis.

Figure 1. A cone generated by revolving the line [latex]y=3x[/latex] around the [latex]y[/latex] -axis.

Conic sections are generated by the intersection of a plane with a cone (Figure 2). If the plane is parallel to the axis of revolution (the y -axis), then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than [latex]90^{\circ}[/latex]), then the conic section is an ellipse.

This figure has three figures. The first figure shows a plain cone with two nappes. The second figure shows a cone with a plane through one nappes and the circle at the top, which creates a parabola. There is also a circle, which occurs when a plane intersects one of the nappes while parallel to the circular bases. There is also an ellipse, which occurs when a plane insects one of the nappes while not parallel to one of the circular bases. Note that the circle and the ellipse are bounded by the edges of the cone on all sides. The last figure shows a hyperbola, which is obtained when a plane intersects both nappes.

Figure 2. The four conic sections. Each conic is determined by the angle the plane makes with the axis of the cone.

Calculus Volume 2. Authored by : Gilbert Strang, Edwin (Jed) Herman. Provided by : OpenStax. Located at : https://openstax.org/books/calculus-volume-2/pages/1-introduction . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike . License Terms : Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction

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High school geometry

Course: high school geometry > unit 7.

Getting ready for conic sections

Intro to conic sections

Graphing circles from features
Graph a circle from its features
Features of a circle from its graph

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Video transcript

Conic Sections

Conic Section: a section (or slice) through a cone .

So all those curves are related.

The curves can also be defined using a straight line (the directrix ) and a point (the focus ).

When we measure the distance:

from the focus to any point on the curve, and
perpendicularly from the directrix to that point

the two distances will always have the same ratio.

For an ellipse, the ratio is less than 1
For a parabola, the ratio is 1 (so the two distances are equal )
For a hyperbola, the ratio is greater than 1

That ratio is called the eccentricity . Play with it here:

Eccentricity

We can say that any conic section is:

"all points whose distance to the focus is equal to the eccentricity times the distance to the directrix "

0 < eccentricity < 1 we get an ellipse,
eccentricity = 1 a parabola, and
eccentricity > 1 a hyperbola.

A circle has an eccentricity of zero , so the eccentricity shows us how "un-circular" the curve is. The bigger the eccentricity, the less curved it is.

Example: Orbits have an eccentricity less than 1

An eccentricity above 1 is is not really an orbit as it does not loop back, but passes by.

The interstellar asteroid 'Oumuamua has an eccentricity of about 1.2 in it's path around the Sun, meaning it is not part of our solar system:

The orbit of Earth has an eccentricity of about 0.0167 (nearly a circle) The orbit of Mars has an eccentricity of about 0.0934 (a little less circular)

Latus Rectum

The latus rectum (no, it is not a rude word!) runs parallel to the directrix and passes through the focus. Its length:

In a parabola, is four times the focal length
In a circle, is the diameter
In an ellipse, is 2b 2 /a (where a and b are one half of the major and minor diameter).

Here is the major axis and minor axis of an ellipse.

There is a focus and directrix on each side (ie a pair of them).

When placed like this on an x-y graph, the equation for an ellipse is:

x 2 a 2 + y 2 b 2 = 1

The special case of a circle (where radius=a=b) is:

x 2 a 2 + y 2 a 2 = 1

And for a hyperbola it is:

x 2 a 2 − y 2 b 2 = 1

General Equation

We can make an equation that covers all these curves.

Because they are plane curves (even though cut out of the solid) we only have to deal with Cartesian ("x" and "y") Coordinates .

But these are not straight lines, so just "x" and "y" will not do ... we need to go to the next level, and have:

x 2 and y 2 ,
and also x (without y), y (without x),
x and y together ( xy )
and a constant term.

There, that should do it!

Give each one a factor (A,B,C etc) and we get a general equation that covers all conic sections:

From that equation we can create equations for the circle, ellipse, parabola and hyperbola.

Conic Section

Conic sections or sections of a cone are the curves obtained by the intersection of a plane and cone. There are three major sections of a cone or conic sections : parabola, hyperbola, and ellipse(the circle is a special kind of ellipse). A cone with two identical nappes is used to produce the conic sections .

All the sections of a cone or conic sections have different shapes but they do share some common properties which we will read in the following sections. Let us check the conic section formulas, conic equations and its parameters, with examples, FAQs.

What Is Conic Section?

Conic sections are the curves obtained when a plane cuts the cone. A cone generally has two identical conical shapes known as nappes. We can get various shapes depending upon the angle of the cut between the plane and the cone and its nappe. By cutting a cone by a plane at different angles, we get the following shapes:

Ellipse is a conic section that is formed when a plane intersects with the cone at an angle. The circle is a special type of ellipse where the cutting plane is parallel to the base of the cone. A hyperbola is formed when the interesting plane is parallel to the axis of the cone, and intersect with both the nappes of the double cone. When the intersecting plane cuts at an angle to the surface of the cone, we get a conic section named parabola.

Conic Section Parameters

The focus, directrix, and eccentricity are the three important features or parameters which defined the conic. The various conic figures are the circle, ellipse, parabola, and hyperbola. And the shape and orientation of these shapes are completely based on these three important features. Let us learn in detail about each of them.

The focus or foci(plural) of a conic section is/are the point(s) about which the conic section is created. They are specially defined for each type of conic section. A parabola has one focus, while ellipses and hyperbolas have two foci. For an ellipse, the sum of the distance of the point on the ellipse from the two foci is constant. Circle, which is a special case of an ellipse, has both the foci at the same place and the distance of all points from the focus is constant. For parabola, it is a limiting case of an ellipse and has one focus at a distance from the vertex, and another focus at infinity. The hyperbola has two foci and the absolute difference of the distance of the point on the hyperbola from the two foci is constant.

Directrix is a line used to define the conic sections. The directrix is a line drawn perpendicular to the axis of the referred conic. Every point on the conic is defined by the ratio of its distance from the directrix and the foci. The directrix is parallel to the conjugate axis and the latus rectum of the conic. A circle has no directrix. The parabola has 1 directrix, the ellipse and the hyperbola have 2 directrices each.

Eccentricity

The eccentricity of a conic section is the constant ratio of the distance of the point on the conic section from the focus and directrix. Eccentricity is used to uniquely define the shape of a conic section. It is a non-negative real number. Eccentricity is denoted by "e". If two conic sections have the same eccentricity, they will be similar. As eccentricity increases, the conic section deviates more and more from the shape of the circle. The value of e for different conic sections is as follows.

For circle, e = 0.
For ellipse, 0 ≤ e < 1
For parabola, e = 1
For hyperbola, e > 1

Terms Related To Conic Section

Other than these three parameters, conic sections have a few more parameters like principal axis, latus rectum, major and minor axis, focal parameter, etc. Let us briefly learn about each of these parameters related to the conic section. The following are the details of the parameters of the conic section.

Principal Axis: The axis passing through the center and foci of a conic is its principal axis and is also referred to as the major axis of the conic.
Conjugate Axis: The axis drawn perpendicular to the principal axis and passing through the center of the conic is the conjugate axis. The conjugate axis is also its minor axis.
Center: The point of intersection of the principal axis and the conjugate axis of the conic is called the center of the conic.
Vertex: The point on the axis where the conic cuts the axis is referred to as the vertex of the conic.
Focal Chord: The focal chord of a conic is the chord passing through the focus of the conic section. The focal chord cuts the conic section at two distinct points.
Focal Distance: The distance of a point $(x_1, y_1)$ on the conic, from any of the foci, is the focal distance. For an ellipse, hyperbola we have two foci, and hence we have two focal distances.
Latus Rectum: It is a focal chord that is perpendicular to the axis of the conic. The length of the latus rectum for a parabola is LL' = 4a. And the length of the latus rectum for an ellipse, and hyperbola is 2b 2 /a.

Tangent: The tangent is a line touching the conic externally at one point on the conic. The point where the tangent touches the conic is called the point of contact. Also from an external point, about two tangents can be drawn to the conic.

Normal: The line drawn perpendicular to the tangent and passing through the point of contact and the focus of the conic is called the normal. We can have one normal for each of the tangents to the conic.

Chord of Contact: The chord drawn to join the point of contact of the tangents, drawn from an external point to the conic is called the chord of contact.

Pole and Polar: For a point which is referred as a pole and lying outside the conic section, the locus of the points of intersection of the tangents, draw at the ends of the chords, drawn from this point is called the polar.

Auxilary Circle: A circle drawn on the major axis of the ellipse as its diameter is called the auxiliary circle. The conic equation of an ellipse is x 2 /a 2 + y 2 /b 2 = 1, and the equation of the auxiliary circle is x 2 + y 2 = a 2 .
Director Circle: The locus of the point of intersection of the perpendicular tangents drawn to the ellipse is called the director circle. For an ellipse (x 2 /a 2 + y 2 /b 2 = 1), the equation of the director circle is x 2 + y 2 = a 2 + b 2

Asymptotes: The pair of straight lines drawn parallel to the hyperbola and assumed to touch the hyperbola at infinity. The equations of the asymptotes of the hyperbola are y = bx/a, and y = -bx/a respectively. And for a hyperbola having the conic equation of x 2 /a 2 - y 2 /b 2 = 1, the equation of the pair of asymptotes of the hyperbola are $\dfrac{x}{a} ± \dfrac{y}{b} = 0$.

Circle - Conic Section

The circle is a special type of ellipse where the cutting plane is parallel to the base of the cone. The circle has a focus known as the center of the circle. The locus of the points on the circle have a fixed distance from the focus or center of the circle and is called the radius of the circle. The value of eccentricity(e) for a circle is e = 0. Circle has no directrix. The general form of the equation of a circle with center at (h, k), and radius r:

(x−h) 2 + (y−k) 2 = r 2

Parabola - Conic Section

When the intersecting plane is at an angle to the surface of the cone we get a conic section named parabola. It is a U-shaped conic section. The value of eccentricity(e) for parabola is e = 1. It is asymmetrical open plane curve formed by the intersection of a cone with a plane parallel to its side. The graph of a quadratic function is a parabola, a line-symmetric curve whose shape is like the graph of y = x 2 . The graph of a parabola either opens upward like y = x 2 or opens downward like the graph of y = - x 2 . The path of a projectile under the influence of gravity ideally follows a curve of this shape.

Ellipse - Conic Section

Ellipse is a conic section that is formed when a plane intersects with the cone at an angle. Ellipse has 2 foci, a major axis, and a minor axis. Value of eccentricity(e) for ellipse is e < 1. Ellipse has 2 directrices. The general form of the equation of an ellipse with center at (h, k) and length of the major and minor axes as '2a' and '2b' respectively. The major axis of the ellipse is parallel to the x-axis. The conic section formula for an ellipse is as follows.

(x−h) 2 /a 2 + (y−k) 2 /b 2 = 1

Note: If the major axis is parallel to the y-axis, switch the places of a and b in the above-given formula.

Hyperbola - Conic Section

A hyperbola is formed when the interesting plane is parallel to the axis of the cone, and intersect with both the nappes of the double cone. The value for eccentricity(e) for hyperbola is e > 1. The two unconnected sections of the hyperbola are called branches. They are mirror images of each other, and their diagonally opposite arms approach the limit to a line.

A hyperbola is an example of a conic section that can be drawn on a plane that intersects a double cone created from two nappes.The general form of the equation of the hyperbola with (h, k) as the center is as follows.

(x−h) 2 /a 2 - (y−k) 2 /b 2 = 1

Conic Section Formulas - Standard Forms

Conic section formulas represent the standard forms of a circle, parabola, ellipse, hyperbola. For ellipses and hyperbolas, the standard form has the x-axis as the principal axis and the origin (0,0) as the center. The vertices are (±a, 0) and the foci (±c, 0)., and is defined by the equations c 2 = a 2 − b 2 for an ellipse and c 2 = a 2 + b 2 for a hyperbola. For a circle, c = 0 so a 2 = b 2 . For the parabola, the standard form has the focus on the x-axis at the point (a, 0) and the directrix is the line with equation x = −a.

Circle: x 2 +y 2 = a 2
Parabola: y 2 = 4ax when a>0
Ellipse: x 2 /a 2 + y 2 /b 2 = 1
Hyperbola: x 2 /a 2 – y 2 /b 2 = 1

Conic Section Examples

Example 1: What will be the equation for the hyperbola which has center at (2, 3), vertex at (0, 3), and the focus at (5, 3).

As we see, for hyperbola, all three points i.e., center, vertex, and focus lie on the same line y = 3.

Now we can see from the given points:

a = 2, c = 3

b 2 = c 2 - a 2 = 9 – 4 = 5.

Putting in the equation of hyperbola conic section:

(x−2) 2 /2 2 - (y−3) 2 /5 = 1

Answer: Equation of the hyperbola will be (x−2) 2 /4 - (y−3) 2 /5 = 1.

Example 2: If for an ellipse, the focus lies at (3, 0), a vertex lies at (4, 0), and its center lies at (0, 0). Find the equation of the ellipse.

From the given points, we can see that

c = 3 and a = 4.

Using b 2 = a 2 – c 2

b 2 = 16 – 9 = 7

Putting in the equation of ellipse conic section:

x 2 /a 2 + y 2 /b 2 = 1

x 2 /16 + y 2 /7 = 1

Answer: The equation of the ellipse is x 2 /16 + y 2 /7 = 1.

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Practice Questions on Conic Section

Faqs on conic section, what is conic section in geometry.

A conic section is a geometric representation of a parabola, ellipse, hyperbola in a two-dimensional coordinate system. These conic are obtained from a simple cone and is obtained by cutting the cone across different sections.

What is Parabola in Conic Section?

When the intersecting plane is at an angle to the surface of the cone, we get a conic section named parabola. It is a U-shaped conic section. The value of eccentricity(e) for parabola is e = 1. It is a symmetrical open plane curve formed by the intersection of a cone with a plane parallel to its side. The path of a projectile under the influence of gravity ideally follows a curve of this shape.

The standard form of the equation of a parabola having the axis along the x-axis, and vertex at the origin is y 2 = 4ax.

What is Circle in Conic Section?

The circle is a special type of ellipse where the cutting plane s parallel to the base of the cone. The circle has a focus known as the center of the circle. The locus of the points on the circle have a fixed distance from the focus or center of the circle and this fixed distance is called the radius of the circle. The value of eccentricity(e) for a circle is e = 0. Circle has no directrix. The general form of the equation of a circle with center at (h, k), and radius r, is as follows.

What is Hyperbola in Conic Section?

A hyperbola is formed when the interesting plane is parallel to the axis of the cone , and intersect with both the nappes of the double cone. The hyperbola represents the locus of a point such that the difference of its distances from the two foci is a constant value. The eccentricity(e) for hyperbola has a value greater than 1. (e > 1)

The general form of the equation of the hyperbola with (h, k) as the center, the x-axis as the major axis, and the y-axis as the minor axis, is as follows.

What is Ellipse in Conic Section?

What is Eccentricity of a Conic Section?

The eccentricity of a conic section is the constant ratio of the distance of the point on the conic section from the focus and directrix. Eccentricity is used to uniquely define the shape of a conic section. It is a non-negative real number, which lies between 0 and 1. The excentricity values for the different conics is as follows.

For hyperbola, e > 1

What are the Applications of the Conic Section?

Here are a few real-life applications of conic sections which we might have seen or known are as follows.

Planets travel around the Sun in elliptical routes at one focus.
Mirrors used to direct light beams at the focus of the parabola are parabolic.
Parabolic mirrors in solar ovens focus light beams for heating.
Sound waves are focused on parabolic microphones.
Car headlights and spotlights are designed based on parabola’s principles.
The path traveled by objects thrown into the air is parabolic.
Hyperbolas are used in long-range navigation systems called LORAN.
Telescopes use parabolic mirrors.

Class 11 Maths
Chapter 11: Conic Sections

Important Questions for Class 11 Maths Chapter 11 - Conic Sections

Important questions for class 11 Maths Chapter 11 – conic sections with solutions are given here. The important questions given here are based on the latest exam pattern and previous year question papers and sample papers. Solving these questions will help the students to score good marks in the annual examinations. The case study questions are framed as per the CBSE board syllabus (2022-2023) and NCERT curriculum . Also, HOTS and value-based questions are asked related to the concept.

Get the Chapter-wise important questions for Class 11 Maths at BYJU’S.

Class 11 Maths Chapter 11 – conic sections will incorporate the concept of sections of cone such as

Learn about these conic sections by solving the questions here. At the end of the article, we have also provided extra questions for chapter 11 conic sections, so that students can practice more on the topic.

Class 11 Chapter 11 – Conic Sections Important Questions with Solutions

To score the good marks in the final examination, practice the problems provided here, which will help you to solve the problems in the annual examination.

Question 1:

Determine the equation of the circle with radius 4 and Centre (-2, 3).

Given that:

Radius, r = 4, and center (h, k) = (-2, 3).

We know that the equation of a circle with centre (h, k) and radius r is given as

(x – h) 2 + (y – k) 2 = r 2 ….(1)

Now, substitute the radius and center values in (1), we get

Therefore, the equation of the circle is

(x + 2) 2 + (y – 3) 2 = (4) 2

x 2 + 4x + 4 + y 2 – 6y + 9 = 16

Now, simplify the above equation, we get:

x 2 + y 2 + 4x – 6y – 3 = 0

Thus, the equation of a circle with center (-2, 3) and radius 4 is x 2 + y 2 + 4x – 6y – 3 = 0

Question 2:

Compute the centre and radius of the circle 2x 2 + 2y 2 – x = 0

Given that, the circle equation is 2x 2 + 2y 2 – x = 0

This can be written as:

⇒ (2x 2 -x) + y 2 = 0

⇒ 2{[x 2 – (x/2)] +y 2 } = 0

⇒{ x 2 – 2x(¼) + (¼) 2 } +y 2 – (¼) 2 = 0

Now, simplify the above form, we get

⇒(x- (¼)) 2 + (y-0) 2 = (¼) 2

The above equation is of the form (x – h) 2 + (y – k) 2 = r 2

Therefore, by comparing the general form and the equation obtained, we can say

h= ¼ , k = 0, and r = ¼.

Question 3:

Determine the focus coordinates, the axis of the parabola, the equation of the directrix and the latus rectum length for y 2 = -8x

Given that, the parabola equation is y 2 = -8x.

It is noted that the coefficient of x is negative.

Therefore, the parabola opens towards the left.

Now, compare the equation with y 2 = -4ax, we obtain

Thus, the value of a is 2.

Therefore, the coordinates of the focus = (-a, 0) = (-2, 0)

Since the given equation involves y 2 , the axis of the parabola is the x-axis.

Equation of directrix, x= a i.e., x = 2

We know the formula to find the length of a latus rectum

Latus rectum length= 4a

Now, substitute a = 2, we get

Length of latus rectum = 8

Question 4:

Determine the foci coordinates, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse (x 2 /49) + (y 2 /36) = 1

The given equation is (x 2 /49) + (y 2 /36) = 1

It can be written as (x 2 /7 2 ) + (y 2 /6 2 ) = 1

It is noticed that the denominator of x 2 /49 is greater than the denominator of the y 2 /36

On comparing the equation with (x 2 /a 2 ) + (y 2 /b 2 ) = 1, we will get

a= 7 and b = 6

Therefore, c = √(a 2 – b 2 )

Now, substitute the value of a and b

⇒ √(a 2 – b 2 ) = √(7 2 – 6 2 ) = √(49-36)

Hence, the foci coordinates are ( ± √13, 0)

Eccentricity, e = c/a = √13/ 7

Length of the major axis = 2a = 2(7) = 14

Length of the minor axis = 2b = 2(6) =12

The coordinates of the vertices are ( ± 7, 0)

Latus rectum Length= 2b 2 /a = 2(6) 2 /7 = 2(36)/7 = 72/7

Question 5:

Determine the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), the major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Centre = (0, 0), and major axis that passes through the points (3, 2) and (1, 6).

We know that the equation of the ellipse will be of the form when the centre is at (0, 0) and the major axis is on the y-axis,

(x 2 /b 2 ) + (y 2 /a 2 ) = 1 …. (1)

Here, a is the semi-major axis.

It is given that, the ellipse passes through the points (3, 2) and (1, 6).

Hence, equation (1) becomes

(9/b 2 ) + (4/a 2 ) = 1 …(2)

(1/b 2 ) + (36/a 2 ) = 1 …(3)

Solving equation (2) and (3), we get

b 2 = 10 and a 2 = 40

Therefore, the equation of the ellipse becomes: (x 2 /10) + (y 2 /40) = 1

Question 6:

Determine the equation of the hyperbola which satisfies the given conditions: Foci (0, ±13), the conjugate axis is of length 24.

Given that: Foci (0, ±13), Conjugate axis length = 24

It is noted that the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form:

(y 2 /a 2 )-(x 2 /b 2 ) = 1 …(1)

Since the foci are (0, ±13), we can get

It is given that, the length of the conjugate axis is 24,

It becomes 2b = 24

And, we know that a 2 + b 2 = c 2

To find a, substitute the value of b and c in the above equation:

a 2 + 12 2 = 13 2

a 2 = 169-144

Now, substitute the value of a and b in equation (1), we get

(y 2 /25)-(x 2 /144) = 1, which is the required equation of the hyperbola.

Practice Problems for Class 11 Maths Chapter 11 Conic Sections

These class 11 Conic Sections questions are categorized into short answer type questions and long answer type questions. These extra questions cover various concepts which will help class 11 students to develop problem-solving skills for the exam.

Calculate the equation of a circle that passes through the origin and cuts off intercepts -2 and 3 from the axis and the y-axis respectively. (Solution: x 2 + y 2 + 2x -3y)
Determine the equation of the circle passing through the points – (0,0)(5,0) and (3,3). (Solution: x 2 + y 2 – 5x -y =0), centre (5/2 , ½) and radius = √ 26/2).
If the distance between the foci of a hyperbola is 16 and eccentricity is √ 2, then obtain its equation. (Solution: x 2 – y 2 =32)
If a latus rectum of an ellipse subtends a right angle at the centre of the ellipse, then write the eccentricity of the ellipse. (Solution: (√ 5 – 1) / 2)
Determine the equation of the ellipse whose foci are (4,0) and (-4,0), eccentricity = ⅓. (Solution: x 2 / 9 + y 2 /8 = 16)
Write the equation of the parabola whose vertex is at (-3,0) and the directrix is (x + 5 ) = 0. (Solution: y 2 = 8(x + 3))
AB is a double ordinate of a parabola y 2 = 4px. Find the locus of its points of trisection. (Solution: 9y 2 =4px)
Calculate the equation of the parabola whose focus is (1, -1) and whose vertex is (2,1). Also, find its axis and latus- rectum). (Solution: 4 √ 5).
Find the equation of the circle which circumscribes the triangle formed by the lines x = 0, y = 0 and lx +my = 1. (Solution: x 2 + y 2 – (1/l)x – (1/m)y = 0)
Prove that the points (9,1) ( 7,9) (-2, 12) and (6,10) are concyclic.
Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and the centre is (0,0).
Find the equation of the circle which touches x-axis and whose centre is (1,2).
Find the coordinates of a point on the parabola y 2 =8x whose focal distance is 4.

For more solved and important problems in Class 11 Maths concepts, register with BYJU’S – The Learning App and download the app today!

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6.5: Conic Sections

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Conic sections have been studied since the time of the ancient Greeks, and were considered to be an important mathematical concept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since then, important applications of conic sections have arisen (for example, in astronomy ), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even architecture. In this section we discuss the three basic conic sections, some of their properties, and their equations.

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Conic sections are generated by the intersection of a plane with a cone (Figure $\PageIndex{2}$). If the plane is parallel to the axis of revolution (the y -axis), then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than 90 ° ) , then the conic section is an ellipse.

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Definitions: The focus, directrix And vertex

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A graph of a typical parabola appears in Figure $\PageIndex{3}$. Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point P with coordinates $(x_1,y_1)$ and point Q with coordinates $(x_2,y_2),$ the distance between them is given by the formula

\[d(P,Q)=\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}.\]

Then from the definition of a parabola and Figure $\PageIndex{3}$, we get

\[d(F,P)=d(P,Q)\]

\[\sqrt{(0−x)^2+(p−y)^2}=\sqrt{(x−x)^2+(−p−y)^2}.\]

Squaring both sides and simplifying yields

\[ \begin{align} x^2+(p−y)^2 &= 0^2+(−p−y)^2 \\ x^2+p^2−2py+y^2 &= p^2+2py+y^2 \\ x^2−2py& =2py \\ x^2& =4py. \end{align}\]

Now suppose we want to relocate the vertex. We use the variables $(h,k)$ to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates $(h,k+p)$ and the directrix has the equation $y=k−p$. Going through the same derivation yields the formula $(x−h)^2=4p(y−k)$. Solving this equation for $y$ leads to the following theorem.

Equations for Parabolas: standard form

Given a parabola opening upward with vertex located at $(h,k)$ and focus located at $(h,k+p)$, where $p$ is a constant, the equation for the parabola is given by

\[y=\dfrac{1}{4p}(x−h)^2+k.\]

This is the standard form of a parabola.

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

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In addition, the equation of a parabola can be written in the general form , though in this form the values of $h$, $k$, and $p$ are not immediately recognizable. The general form of a parabola is written as

\[ax^2+bx+cy+d=0 \label{para1}\]

\[ay^2+bx+cy+d=0.\label{para2}\]

Equation \ref{para1} represents a parabola that opens either up or down. Equation \ref{para2} represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.

Example $\PageIndex{1}$: Converting the Equation of a Parabola from General into Standard Form

Put the equation

\[x^2−4x−8y+12=0\]

into standard form and graph the resulting parabola.

\[8y=x^2−4x+12.\]

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:

\[8y=(x^2−4x)+12.\]

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of x and square it. This gives $(\dfrac{−4}{2})^2=4.$ Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

\[8y=(x^2−4x+4)+12−4.\]

Now combine like terms and factor the quantity inside the parentheses:

\[8y=(x−2)^2+8.\]

Finally, divide by 8:

\[y=\dfrac{1}{8}(x−2)^2+1.\]

This equation is now in standard form. Comparing this to Equation gives $h=2, k=1$, and $p=2$. The parabola opens up, with vertex at $(2,1)$, focus at $(2,3)$, and directrix $y=−1$. The graph of this parabola appears as follows.

$11_5_1.png$

Exercise $\PageIndex{1}$

Put the equation $2y^2−x+12y+16=0$ into standard form and graph the resulting parabola.

Solve for $x$. Check which direction the parabola opens.

\[x=2(y+3)^2−2\]

$11_5_2.png$

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Definition: Ellipse

An ellipse is the set of all points for which the sum of their distances from two fixed points (the foci) is constant.

A graph of a typical ellipse is shown in Figure $\PageIndex{6}$. In this figure the foci are labeled as $F$ and $F′$. Both are the same fixed distance from the origin, and this distance is represented by the variable c. Therefore the coordinates of $F$ are $(c,0)$ and the coordinates of $F′$ are $(−c,0).$ The points $P$ and $P′$ are located at the ends of the major axis of the ellipse, and have coordinates $(a,0)$ and $(−a,0)$, respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2a. Furthermore, $P$ and $P′$ are called the vertices of the ellipse. The points $Q$ and $Q′$ are located at the ends of the minor axis of the ellipse, and have coordinates $(0,b)$ and $(0,−b),$ respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.

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\[d(P,F)+d(P,F′)=(a−c)+(a+c)=2a.\]

Therefore the sum of the distances from an arbitrary point A with coordinates $(x,y)$ is also equal to $2a$. Using the distance formula, we get

\[d(A,F)+d(A,F′)=2a.\]

\[\sqrt{(x−c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a\]

Subtract the second radical from both sides and square both sides:

\[\sqrt{(x−c)^2+y^2}=2a−\sqrt{(x+c)^2+y^2}\]

\[(x−c)^2+y^2=4a^2−4a\sqrt{(x+c)^2+y^2}+(x+c)^2+y^2\]

\[x^2−2cx+c^2+y^2=4a^2−4a\sqrt{(x+c)^2+y^2}+x^2+2cx+c^2+y^2\]

\[−2cx=4a^2−4a\sqrt{(x+c)^2+y^2}+2cx.\]

Now isolate the radical on the right-hand side and square again:

\[−2cx=4a^2−4a\sqrt{(x+c)^2+y^2}+2cx\]

\[4a\sqrt{(x+c)^2+y^2}=4a^2+4cx\]

\[\sqrt{(x+c)^2+y^2}=a+\dfrac{cx}{a}\]

\[(x+c)^2+y^2=a^2+2cx+\dfrac{c^2x^2}{a^2}\]

\[x^2+2cx+c^2+y^2=a^2+2cx+\dfrac{c^2x^2}{a^2}\]

\[x^2+c^2+y^2=a^2+\dfrac{c^2x^2}{a^2}.\]

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

\[x^2−\dfrac{c^2x^2}{a^2}+y^2=a^2−c^2\]

\[\dfrac{(a^2−c^2)x^2}{a^2}+y^2=a^2−c^2.\]

Divide both sides by $a^2−c^2$. This gives the equation

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2−c^2}=1.\]

If we refer back to Figure $\PageIndex{6}$, then the length of each of the two green line segments is equal to $a$. This is true because the sum of the distances from the point Q to the foci $F$ and $F′$ is equal to $2a$, and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length a and leg lengths b and c. From the Pythagorean theorem, $a^2+b^2=c^2$ and $b^2=a^2−c^2$. Therefore the equation of the ellipse becomes

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.\]

Finally, if the center of the ellipse is moved from the origin to a point $(h,k)$, we have the following standard form of an ellipse.

Equation of an Ellipse in Standard Form

Consider the ellipse with center $(h,k)$, a horizontal major axis with length $2a$, and a vertical minor axis with length $2b$. Then the equation of this ellipse in standard form is

\[\dfrac{(x−h)^2}{a^2}+\dfrac{(y−k)^2}{b^2}=1\]

and the foci are located at $(h±c,k)$, where $c^2=a^2−b^2$. The equations of the directrices are $x=h±\dfrac{a^2}{c}$.

If the major axis is vertical, then the equation of the ellipse becomes

\[\dfrac{(x−h)^2}{b^2}+\dfrac{(y−k)^2}{a^2}=1\]

and the foci are located at $(h,k±c)$, where $c^2=a^2−b^2$. The equations of the directrices in this case are $y=k±\dfrac{a^2}{c}$.

\[(Ax^2+By^2+Cx+Dy+E=0,\]

where A and B are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square .

Example $\PageIndex{2}$: Finding the Standard Form of an Ellipse

\[9x^2+4y^2−36x+24y+36=0\]

into standard form and graph the resulting ellipse.

First subtract 36 from both sides of the equation:

\[9x^2+4y^2−36x+24y=−36.\]

Next group the $x$ terms together and the $y$ terms together, and factor out the common factor:

\[(9x^2−36x)+(4y^2+24y)=−36\]

\[9(x^2−4x)+4(y^2+6y)=−36.\]

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives $(\dfrac{−4}{2})^2=4.$ In the second set of parentheses, take half the coefficient of y and square it. This gives $(\dfrac{6}{2})^2=9.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

\[9(x^2−4x+4)+4(y^2+6y+9)=−36+36+36\]

\[9(x^2−4x+4)+4(y^2+6y+9)=36.\]

Now factor both sets of parentheses and divide by 36:

\[9(x−2)^2+4(y+3)^2=36\]

\[\dfrac{9(x−2)^2}{36}+\dfrac{4(y+3)^2}{36}=1\]

\[\dfrac{(x−2)^2}{4}+\dfrac{(y+3)^2}{9}=1.\]

The equation is now in standard form. Comparing this to Equation gives $h=2, k=−3, a=3,$ and $b=2$. This is a vertical ellipse with center at $(2,−3)$, major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

$11_5_3.png$

Exercise $\PageIndex{2}$

\[9x^2+16y^2+18x−64y−71=0\]

Move the constant over and complete the square.

\[\dfrac{(x+1)^2}{16}+\dfrac{(y−2)^2}{9}=1\]

$11_5_4.png$

According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in Figure $\PageIndex{8A}$. Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth.

Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focus after mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure $\PageIndex{8B}$. This hall served as the meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they can hear someone standing close by. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Central Station in New York City.

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A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.

Definition: hyperbola

A hyperbola is the set of all points where the difference between their distances from two fixed points (the foci) is constant.

A graph of a typical hyperbola appears as follows.

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The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let P be a point on the hyperbola with coordinates $(x,y)$. Then the definition of the hyperbola gives $|d(P,F_1)−d(P,F_2)|=constant$. To simplify the derivation, assume that P is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates $(a,0),$ so

\[d(P,F_1)−d(P,F_2)=(c+a)−(c−a)=2a.\]

This equation is therefore true for any point on the hyperbola. Returning to the coordinates $(x,y)$ for P :

\[d(P,F_1)−d(P,F_2)=2a\]

\[\sqrt{(x+c)^2+y^2}−\sqrt{(x−c)^2+y^2}=2a.\]

Add the second radical from both sides and square both sides:

\[\sqrt{(x−c)^2+y^2}=2a+\sqrt{(x+c)^2+y^2}\]

\[(x−c)^2+y2=4a^2+4a\sqrt{(x+c)^2+y^2}+(x+c)^2+y^2\]

\[x^2−2cx+c^2+y^2=4a^2+4a\sqrt{(x+c)^2+y^2}+x^2+2cx+c^2+y^2\]

\[−2cx=4a^2+4a\sqrt{(x+c)^2+y^2}+2cx.\]

$−2cx=4a^2+4a\sqrt{(x+c)^2+y^2}+2cx$

$4a\sqrt{(x+c)^2+y^2}=−4a^2−4cx$

$(x+c)^2+y^2=−a−\dfrac{cx}{a}$

$(x+c)^2+y^2=a^2+2cx+\dfrac{c^2x^2}{a^2}$

$x^2+2cx+c^2+y^2=a^2+2cx+\dfrac{c^2x^2}{a^2}$

$x^2+c^2+y^2=a^2+\dfrac{c^2x^2}{a^2}$.

Finally, divide both sides by $a^2−c^2$. This gives the equation

We now define b so that $b^2=c^2−a^2$. This is possible because $c>a$. Therefore the equation of the ellipse becomes

\[\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1.\]

Finally, if the center of the hyperbola is moved from the origin to the point $(h,k),$ we have the following standard form of a hyperbola.

Equation of a Hyperbola in Standard Form

Consider the hyperbola with center $(h,k)$, a horizontal major axis, and a vertical minor axis. Then the equation of this ellipse is

\[\dfrac{(x−h)^2}{a^2}−\dfrac{(y−k)^2}{b^2}=1\]

and the foci are located at $(h±c,k),$ where $c^2=a^2+b^2$. The equations of the asymptotes are given by $y=k±\dfrac{b}{a}(x−h).$ The equations of the directrices are

\[x=k±\dfrac{a^2}{\sqrt{a^2+b^2}}=h±\dfrac{a^2}{c}\]

If the major axis is vertical, then the equation of the hyperbola becomes

\[\dfrac{(y−k)^2}{a^2}−\dfrac{(x−h)^2}{b^2}=1\]

and the foci are located at $(h,k±c),$ where $c^2=a^2+b^2$. The equations of the asymptotes are given by $y=k±\dfrac{a}{b}(x−h)$. The equations of the directrices are

\[y=k±\dfrac{a^2}{\sqrt{a^2+b^2}}=k±\dfrac{a^2}{c}.\]

\[Ax^2+By^2+Cx+Dy+E=0,\]

where A and B have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

Example $\PageIndex{3}$: Finding the Standard Form of a Hyperbola

Put the equation $9x^2−16y^2+36x+32y−124=0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

First add 124 to both sides of the equation:

$9x^2−16y^2+36x+32y=124.$

Next group the x terms together and the y terms together, then factor out the common factors:

$(9x^2+36x)−(16y^2−32y)=124$

$9(x^2+4x)−16(y^2−2y)=124$.

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives $(\dfrac{4}{2})^2=4$. In the second set of parentheses, take half the coefficient of y and square it. This gives $(\dfrac{−2}{2})^2=1.$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

$9(x^2+4x+4)−16(y^2−2y+1)=124+36−16$

$9(x^2+4x+4)−16(y^2−2y+1)=144.$

Next factor both sets of parentheses and divide by 144:

$9(x+2)^2−16(y−1)^2=144$

$\dfrac{9(x+2)^2}{144}−\dfrac{16(y−1)^2}{144}=1$

$\dfrac{(x+2)^2}{16}−\dfrac{(y−1)^2}{9}=1.$

The equation is now in standard form. Comparing this to Equation gives $h=−2, k=1, a=4,$ and $b=3$. This is a horizontal hyperbola with center at $(−2,1)$ and asymptotes given by the equations $y=1±\dfrac{3}{4}(x+2)$. The graph of this hyperbola appears in Figure $\PageIndex{10}$.

$11_5_5.png$

Exercise $\PageIndex{3}$

Put the equation $4y^2−9x^2+16y+18x−29=0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Move the constant over and complete the square. Check which direction the hyperbola opens

$\dfrac{(y+2)^2}{9}−\dfrac{(x−1)^2}{4}=1.$ This is a vertical hyperbola. Asymptotes $y=−2±\dfrac{3}{2}(x−1).$

$11_5_6.png$

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Eccentricity and Directrix

Definition: Eccentricity and Directrices

The eccentricity $e$ of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix. This value is constant for any conic section, and can define the conic section as well:

If $e=1$, the conic is a parabola.
If $e<1$, it is an ellipse.
If $e>1,$ it is a hyperbola.

The three conic sections with their directrices appear in Figure $\PageIndex{12}$.

$alt$

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are $x=±\dfrac{a^2}{c}$. The right vertex of the ellipse is located at $(a,0)$ and the right focus is $(c,0)$. Therefore the distance from the vertex to the focus is $a−c$ and the distance from the vertex to the right directrix is $\dfrac{a^2}{c}−c.$ This gives the eccentricity as

\[e=\dfrac{a−c}{\dfrac{a^2}{c}−a}=\dfrac{c(a−c)}{a^2−ac}=\dfrac{c(a−c)}{a(a−c)}=\dfrac{c}{a}.\]

Since $c<a$, this step proves that the eccentricity of an ellipse is less than 1. The directrices of a horizontal hyperbola are also located at $x=±\dfrac{a^2}{c}$, and a similar calculation shows that the eccentricity of a hyperbola is also e=ca. However in this case we have $c>a$, so the eccentricity of a hyperbola is greater than 1.

Example $\PageIndex{4}$: Determining Eccentricity of a Conic Section

Determine the eccentricity of the ellipse described by the equation

$\dfrac{(x−3)^2}{16}+\dfrac{(y+2)^2}{25}=1.$

From the equation we see that $a=5$ and $b=4$. The value of c can be calculated using the equation $a^2=b^2+c^2$ for an ellipse. Substituting the values of a and b and solving for c gives $c=3$. Therefore the eccentricity of the ellipse is $e=\dfrac{c}{a}=\dfrac{3}{5}=0.6.$

Exercise $\PageIndex{4}$

Determine the eccentricity of the hyperbola described by the equation

$\dfrac{(y−3)^2}{49}−\dfrac{(x+2)^2}{25}=1.$

First find the values of a and b, then determine c using the equation $c^2=a^2+b^2$.

$e=\dfrac{c}{a}=\dfrac{\sqrt{74}}{7}≈1.229$

Polar Equations of Conic Sections

Polar Equation of Conic Sections

The polar equation of a conic section with focal parameter p is given by

$r=\dfrac{ep}{1±e\cos θ}$ or $r=\dfrac{ep}{1±e\sin θ}.$

Example $\PageIndex{5}$: Graphing a Conic Section in Polar Coordinates

Identify and create a graph of the conic section described by the equation

$r=\dfrac{3}{1+2\cos θ}$.

The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter p can be calculated by using the equation $ep=3.$ Since $e=2$, this gives $p=\dfrac{3}{2}$. The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for $θ$ and create a table of values. Then we can graph the hyperbola (Figure $\PageIndex{13}$).

$11_5_7.png$

Exercise $\PageIndex{5}$

$r=\dfrac{4}{1−0.8 \sin θ}$.

First find the values of e and p , and then create a table of values.

Here $e=0.8$ and $p=5$. This conic section is an ellipse.

$alt$

General Equations of Degree Two

A general equation of degree two can be written in the form

\[ Ax^2+Bxy+Cy^2+Dx+Ey+F=0.\]

The graph of an equation of this form is a conic section. If $B≠0$ then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section $4AC−B^2.$

Identifying the Conic Section

One of the following cases must be true:

$4AC−B^2>0$. If so, the graph is an ellipse.
$4AC−B^2=0$. If so, the graph is a parabola.
$4AC−B^2<0$. If so, the graph is a hyperbola.

The simplest example of a second-degree equation involving a cross term is $xy=1$. This equation can be solved for $y$ to obtain $y=\dfrac{1}{x}$. The graph of this function is called a rectangular hyperbola as shown.

$alt$

The asymptotes of this hyperbola are the $x$ and $y$ coordinate axes. To determine the angle θ of rotation of the conic section, we use the formula $\cot 2θ=\frac{A−C}{B}$. In this case $A=C=0$ and $B=1$, so $\cot 2θ=(0−0)/1=0$ and $θ=45°$. The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled $A′,B′,C′,D′,E′,$ and $F′,$ and are given by the formulas

\[ \begin{align} A′ &=A\cos^ 2θ+B\cos θ\sin θ+C\sin^2 θ \\ B′&=0 \\ C′&=A\sin^2 θ−B\sin θ\cos θ+C\cos^2θ \\ D′&=D\cos θ+E\sin θ \\ E′&=−D\sin θ+E\cosθ \\ F′&=F. \end{align}\]

Procedure: graphing a rotated conic

The procedure for graphing a rotated conic is the following:

Identify the conic section using the discriminant $4AC−B^2$.
Determine $θ$ using the formula \[\cot2θ=\dfrac{A−C}{B} \label{rot}.\]
Calculate $A′,B′,C′,D′,E′$,and $F′$.
Rewrite the original equation using $A′,B′,C′,D′,E′$,and $F′$.
Draw a graph using the rotated equation.

Example $\PageIndex{6}$: Identifying a Rotated Conic

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

\[13x^2−6\sqrt{3}xy+7y^2−256=0.\]

In this equation, $A=13,B=−6\sqrt{3},C=7,D=0,E=0,$ and $F=−256$. The discriminant of this equation is

\[4AC−B^2=4(13)(7)−(−6\sqrt{3})^2=364−108=256.\]

Therefore this conic is an ellipse.

To calculate the angle of rotation of the axes, use Equation \ref{rot}

\[\cot 2θ=\dfrac{A−C}{B}.\]

$\cot 2θ=\dfrac{A−C}{B}=\dfrac{13−7}{−6\sqrt{3}}=−\dfrac{\sqrt{3}}{3}$.

Therefore $2θ=120^o$ and $θ=60^o$, which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

$A′=A\cos^2θ+B\cos θ\sinθ+C\sin^2θ$

$=13\cos^260+(−6\sqrt{3})\cos 60 \sin 60+7\sin^260$

$=13(\dfrac{1}{2})^2−6\sqrt{3}(\dfrac{1}{2})(\dfrac{\sqrt{3}}{2})+7(\dfrac{\sqrt{3}}{2})^2$

$B′=0$

$C′=A\sin^2θ−B\sin θ\cos θ+C\cos^2θ$

$=13\sin^260+(−6\sqrt{3})\sin 60 \cos 60=7\cos^260$

$=(\dfrac{\sqrt{3}}{2})^2+6\sqrt{3}(\dfrac{\sqrt{3}}{2})(\dfrac{1}{2})+7(\dfrac{1}{2})^2$

$D′=Dcosθ+Esinθ$

$=(0)cos60+(0)sin60$

$E′=−Dsinθ+Ecosθ$

$=−(0)sin60+(0)cos60$

$F′= F$

$=−256.$

The equation of the conic in the rotated coordinate system becomes

$4(x′)^2+16(y′)^2=256$

$\dfrac{(x′)^2}{64}+\dfrac{(y′)^2}{16}=1$.

A graph of this conic section appears as follows.

$11_5_8.png$

Exercise $\PageIndex{6}$

\[3x^2+5xy−2y^2−125=0.\]

Follow steps 1 and 2 of the five-step method outlined above

The conic is a hyperbola and the angle of rotation of the axes is $θ=22.5°.$

Key Concepts

The equation of a vertical parabola in standard form with given focus and directrix is $y=\dfrac{1}{4p}(x−h)^2+k$ where $p$ is the distance from the vertex to the focus and $(h,k)$ are the coordinates of the vertex.
The equation of a horizontal ellipse in standard form is $\dfrac{(x−h)^2}{a^2}+\dfrac{(y−k)^2}{b^2}=1$ where the center has coordinates $(h,k)$, the major axis has length 2a , the minor axis has length 2b , and the coordinates of the foci are $(h±c,k)$, where $c^2=a^2−b^2$.
The equation of a horizontal hyperbola in standard form is $\dfrac{(x−h)^2}{a^2}−\dfrac{(y−k)^2}{b^2}=1$ where the center has coordinates $(h,k)$, the vertices are located at $(h±a,k)$, and the coordinates of the foci are $(h±c,k),$ where $c^2=a^2+b^2$.
The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
The polar equation of a conic section with eccentricity e is $r=\dfrac{ep}{1±ecosθ}$ or $r=\dfrac{ep}{1±esinθ}$, where p represents the focal parameter.
To identify a conic generated by the equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$,first calculate the discriminant $D=4AC−B^2$. If $D>0$ then the conic is an ellipse, if $D=0$ then the conic is a parabola, and if $D<0$ then the conic is a hyperbola.

Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .

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Introduction. Conics or conic sections were studied by Greek mathematicians, with Apollonius of Pergo's work on their properties around 200 B.C. Conics sections are planes, cut at varied angles from a cone. The shapes vary according to the angle at which it is cut from the cone. As they are cut from cones, they are called Conies.
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A conic section (or simply conic) is a curve obtained as the intersection of the surface of a cone with a plane. The three types of conic sections are the hyperbola, the parabola, and the ellipse. The circle is type of ellipse, and is sometimes considered to be a fourth type of conic section. Conic sections can be generated by intersecting a ...
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Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions. Level UP. After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS. These CBSE Question Bank for Class 11 Maths Chapter 11 Conic Sections are latest, comprehensive, confidence inspiring, with easy to understand solutions.

Conic Sections

Study Guides > Boundless Algebra

Defining Conic Sections

Common Parts of Conic Sections

Applications of Conic Sections

Defining Eccentricity

Conceptualizing Eccentricity

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7.5 Conic Sections

Theorem 7.8

Example 7.19

Checkpoint 7.18

Theorem 7.9

Example 7.20

Checkpoint 7.19

Theorem 7.10

Example 7.21

Checkpoint 7.20

Eccentricity and Directrix

Example 7.22

Checkpoint 7.21

Polar Equations of Conic Sections

Theorem 7.11

Example 7.23

Checkpoint 7.22

Example 7.24

Checkpoint 7.23

Module 7: Parametric Equations and Polar Coordinates

High school geometry

Intro to conic sections

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Video transcript

Conic Sections

Eccentricity

Example: Orbits have an eccentricity less than 1

Latus Rectum

General Equation

Conic Section

What Is Conic Section?

Conic Section Parameters

Eccentricity

Terms Related To Conic Section

Circle - Conic Section

Parabola - Conic Section

Ellipse - Conic Section

Hyperbola - Conic Section

Conic Section Formulas - Standard Forms

Conic Section Examples

Practice Questions on Conic Section

What is Parabola in Conic Section?

What is Circle in Conic Section?

What is Hyperbola in Conic Section?

What is Ellipse in Conic Section?

What is Eccentricity of a Conic Section?

What are the Applications of the Conic Section?

Important Questions for Class 11 Maths Chapter 11 - Conic Sections

Class 11 Chapter 11 – Conic Sections Important Questions with Solutions

More Articles for Class 11

Practice Problems for Class 11 Maths Chapter 11 Conic Sections

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6.5: Conic Sections

Eccentricity and Directrix

Polar Equations of Conic Sections

General Equations of Degree Two

Key Concepts

Contributors and Attributions

CBSE Question Bank for Class 11 Maths Chapter 11 Conic Sections Free PDF

CBSE Question Bank for Class 11 Maths Chapter 11 Conic Sections PDF

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