case study on conic sections class 11

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Class 11 Mathematics Case Study Questions

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If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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Chapter 10 Class 11 Conic Sections

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Updated for new NCERT Book - 2023-24 .

Learn Chapter 10 Conic Sections of Class 11 free with solutions of all NCERT Questions, Examples and Miscellaneous exercises. All solutions are provided with step-by-step explanation for your reference.

Let's see what conic section is.

We learned Straight Lines in the last chapter, but straight lines are not the only type of curves we have.

In this chapter, we talk about Conic Sections,

that is, sections of the cone

Specifically, we talk about 

Circles, Ellipse, Parabola and Hyperbola

So, the topics of the chapter include

• Circles - How to find equation of circle, center of circle
• Parabola - Equation of parabola, its directrix, eccentricity and focus
• Ellipse - Equation of ellipse, its directrix, eccentricity, focus and vertices
• Hyperbola - Equation of hyperbola, its directrix, eccentricity, focus and vertices
• Other questions like mirror problem , Triangle in parabola problem, Beam problem, Locus, Path traced problems

Click on an exercise to start with answers of the questions, or the topic to learn the concepts with the questions

Serial order wise

Concept wise.

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Unit 10: Conic sections

Introduction to conic sections.

• Intro to conic sections (Opens a modal)

The features of a circle

• Graphing circles from features (Opens a modal)
• Features of a circle from its graph (Opens a modal)
• Graph a circle from its features Get 3 of 4 questions to level up!
• Features of a circle from its graph Get 3 of 4 questions to level up!

Standard equation of a circle

• Features of a circle from its standard equation (Opens a modal)
• Graphing a circle from its standard equation (Opens a modal)
• Writing standard equation of a circle (Opens a modal)
• Features of a circle from its standard equation Get 3 of 4 questions to level up!
• Graph a circle from its standard equation Get 3 of 4 questions to level up!
• Write standard equation of a circle Get 3 of 4 questions to level up!

Expanded equation of a circle

• Features of a circle from its expanded equation (Opens a modal)
• Circle equation review (Opens a modal)
• Features of a circle from its expanded equation Get 3 of 4 questions to level up!
• Graph a circle from its expanded equation Get 3 of 4 questions to level up!

Center and radii of an ellipse

• Intro to ellipses (Opens a modal)
• Ellipse standard equation from graph (Opens a modal)
• Ellipse graph from standard equation (Opens a modal)
• Ellipse features review (Opens a modal)
• Ellipse equation review (Opens a modal)
• Graph & features of ellipses Get 3 of 4 questions to level up!
• Center & radii of ellipses from equation Get 3 of 4 questions to level up!
• Ellipse standard equation & graph Get 3 of 4 questions to level up!

Foci of an ellipse

• Foci of an ellipse from equation (Opens a modal)
• Ellipse foci review (Opens a modal)
• Foci of an ellipse from radii Get 3 of 4 questions to level up!
• Foci of an ellipse from equation Get 3 of 4 questions to level up!
• Equation of an ellipse from features Get 3 of 4 questions to level up!

Focus and directrix of a parabola

• Intro to focus & directrix (Opens a modal)
• Equation of a parabola from focus & directrix (Opens a modal)
• Focus & directrix of a parabola from equation (Opens a modal)
• Parabola focus & directrix review (Opens a modal)
• Equation of a parabola from focus & directrix Get 3 of 4 questions to level up!

Introduction to hyperbolas

• Intro to hyperbolas (Opens a modal)
• Vertices & direction of a hyperbola (Opens a modal)
• Vertices & direction of a hyperbola (example 2) (Opens a modal)
• Graphing hyperbolas (old example) (Opens a modal)
• Vertices & direction of a hyperbola Get 3 of 4 questions to level up!

Hyperbolas not centered at the origin

• Equation of a hyperbola not centered at the origin (Opens a modal)

Identifying conic sections from their equation

• Conic section from expanded equation: circle & parabola (Opens a modal)
• Conic section from expanded equation: ellipse (Opens a modal)
• Conic section from expanded equation: hyperbola (Opens a modal)

CBSE Class 11 Maths – Chapter 11 Conic Sections- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Conic Sections : Notes and Study Materials -pdf

• Concepts of  Conic Sections
• Conic Sections Master File
• R D Sharma Solution of Parabola
• R D Sharma Solution of Hyperbola
• R D Sharma Solution of Ellipse
• R D Sharma Solution of Circle
• NCERT Solution  Conic Sections
• NCERT  Exemplar Solution Conic Sections
• Conic Sections : Solved Example 1

CBSE Class 11 Maths Notes Chapter 11 Conic Sections

Circle A circle is the set of all points in a plane, which are at a fixed distance from a fixed point in the plane. The fixed point is called the centre of the circle and the distance from centre to any point on the circle is called the radius of the circle. The equation of a circle with radius r having centre (h, k) is given by (x – h) 2  + (y – k) 2  = r 2 .

The general equation of the circle is given by x 2  + y 2  + 2gx + 2fy + c = 0 , where, g, f and c are constants.

• The centre of the circle is (-g, -f).
• The radius of the circle is r =  g 2 + f 2 − c −−−−−−−−−√

The general equation of the circle passing through origin is x 2  + y 2  + 2gx + 2fy = 0.

The parametric equation of the circle x 2  + y 2  = r 2  are given by x = r cos θ, y = r sin θ, where θ is the parametre and the parametric equation of the circle (x – h) 2  + (y – k) 2  = r 2  are given by x = h + r cos θ, y = k + r sin θ.

Note: The general equation of the circle involves three constants which implies that at least three conditions are required to determine a circle uniquely.

Parabola A parabola is the set of points P whose distances from a fixed point F in the plane are equal to their distance from a fixed line l in the plane. The fixed point F is called focus and the fixed line l is the directrix of the parabola.

Main Facts About the Parabola

Conic Sections Class 11 MCQs Questions with Answers

Question 1. The locus of the point from which the tangent to the circles x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0 are equal is given by the equation (a) 8x + 19 = 0 (b) 8x – 19 = 0 (c) 4x – 19 = 0 (d) 4x + 19 = 0

Answer: (b) 8x – 19 = 0 Hint: Given equation of circles are x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0 Now, the required line is the radical axis of the two circles are (x² + y² – 4) – (x² + y² – 8x + 15) = 0 ⇒ x² + y² – 4 – x² – y² + 8x – 15 = 0 ⇒ 8x – 19 = 0

Question 2. The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0 (a) 7 (b) 8 (c) 9 (d) 10

Answer: (a) 7 Hint: The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²) = {9 + 16 + 10}/√(9 + 16) = 35/√25 = 35/5 = 7

Question 3. A man running a race course notes that the sum of the distances from the two flag posts from him is always 10 meter and the distance between the flag posts is 8 meter. The equation of posts traced by the man is (a) x²/9 + y²/5 = 1 (b) x²/9 + y2 /25 = 1 (c) x²/5 + y²/9 = 1 (d) x²/25 + y²/9 = 1

Question 4. The center of the ellipse (x + y – 2)² /9 + (x – y)² /16 = 1 is (a) (0, 0) (b) (0, 1) (c) (1, 0) (d) (1, 1)

Answer: (d) (1, 1) Hint: The center of the given ellipse is the point of intersection of the lines x + y – 2 = 0 and x – y = 0 After solving, we get x = 1, y = 1 So, the center of the ellipse is (1, 1)

Question 5. The parametric coordinate of any point of the parabola y² = 4ax is (a) (-at², -2at) (b) (-at², 2at) (c) (a sin²t, -2a sin t) (d) (a sin t, -2a sin t)

Answer: (c) (a sin²t, -2a sin t) Hint: The point (a sin²t, -2a sin t) satisfies the equation of the parabola y² = 4ax for all values of t. So, the parametric coordinate of any point of the parabola y² = 4ax is (a sin²t, -2a sin t)

Question 6. The equation of parabola with vertex at origin the axis is along x-axis and passing through the point (2, 3) is (a) y² = 9x (b) y² = 9x/2 (c) y² = 2x (b) y² = 2x/9

Answer: (b) y² = 9x/2 Hint: A parabola with its axis along the x-axis and vertex(0, 0) and direction x = -a has the equation: y² = 4ax ………….. 1 Given, point (2,3) lies on the parabola, ⇒ 3² = 4a × 2 ⇒ 9 = 4a × 2 ⇒ 9/2 = 4a From equation 1, we get y² = (9/2)x ⇒ y² = 9x/2 This is the required equation of the parabola.

Question 7. At what point of the parabola x² = 9y is the abscissa three times that of ordinate (a) (1, 1) (b) (3, 1) (c) (-3, 1) (d) (-3, -3)

Answer: (b) (3, 1) Hint: Given, parabola is x² = 9y Let P(h, k) is the point on the parabola such that abscissa is 3 times the ordinate. So, h = 3k ……… 1 Since P(h, k) lies on the parabola So, h² = 9k ……… 2 From equation 1 and 2, we get (3k)² = 9k ⇒ 9k² = 9k ⇒ 9k² – 9k = 0 ⇒ 9k(k – 1) = 0 ⇒ k = 0, 1 When k = 0, h = 0 So k = 1 Now, from equation 1, h = 3 × 1 = 3 So, the point is (3, 1)

Question 8. The number of tangents that can be drawn from (1, 2) to x² + y² = 5 is (a) 0 (b) 1 (c) 2 (d) More than 2

Answer: (b) 1 Hint: Given point (1, 2) and equation of circle is x² + y² = 5 Now, x² + y² – 5 = 0 Put (1, 2) in this equation, we get 1² + 2² – 5 = 1 + 4 – 5 = 5 – 5 = 0 So, the point (1, 2) lies on the circle. Hence, only one tangent can be drawn.

Question 9. In an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is (a) 4/5 (b) 1/√52 (c) 3/5 (d) 1/2

Answer: (c) 3/5 Hint: Given, distance between foci = 6 ⇒ 2ae = 6 ⇒ ae = 3 Again minor axis = 8 ⇒ 2b = 8 ⇒ b = 4 ⇒ b² = 16 ⇒ a² (1 – e²) = 16 ⇒ a² – a² e² = 16 ⇒ a² – (ae)² = 16 ⇒ a² – 3² = 16 ⇒ a² – 9 = 16 ⇒ a² = 9 + 16 ⇒ a² = 25 ⇒ a = 5 Now, ae = 3 ⇒ 5e = 3 ⇒ e = 3/5 So, the eccentricity is 3/5

Question 10. If the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is (a) (x + 2)² + (y – 3)² = 3² (b) (x – 2)² + (y + 3)² = 3² (c) (x – 2)² + (y – 3)² = 3² (d) (x + 2)² + (y + 3)² = 3²

Answer: (c) (x – 2)² + (y – 3)² = 3² Hint: Radius of the circle = √{(2 – 0)² + (3 – 0)² – 2²} = √(4 + 9 – 4) = √9 = 3 So, the equation of the circle = (x – 2)² + (y – 3)² = 3²

Question 11. The equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is (a) 16x² – 9y² – 24xy – 144x + 8y + 224 = 0 (b) 16x² + 9y² – 24xy – 144x + 8y – 224 = 0 (c) 16x² + 9y² – 24xy – 144x – 8y + 224 = 0 (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0 Hint: Given focus S(3, 0) and equation of directrix is: 3x + 4y = 1 ⇒ 3x + 4y – 1 = 0 Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix Then, SP = PM ⇒ SP² = PM² ⇒ (x – 3)² + (y – 0)² = {(3x + 4y – 1) /{√(3² + 4²)}² ⇒ x² + 9 – 6x + y² = (9x² + 16y² + 1 + 24xy – 8y – 6x)/25 ⇒ 25(x² + 9 – 6x + y²) = 9x² + 16y² + 1 + 24xy – 8y – 6x ⇒ 25x² + 225 – 150x + 25y² = 9x² + 16y² + 1 + 24xy – 8y – 6x ⇒ 25x² + 225 – 150x + 25y² – 9x² – 16y² – 1 – 24xy + 8y + 6x = 0 ⇒ 16x² + 9y² – 24xy – 144x + 8y + 224 = 0 This is the required equation of parabola.

Question 12. The parametric representation (2 + t², 2t + 1) represents (a) a parabola (b) a hyperbola (c) an ellipse (d) a circle

Answer: (a) a parabola Hint: Let x = 2 + t² ⇒ x – 2 = t² ……….. 1 and y = 2t + 1 ⇒ y – 1 = 2t ⇒ (y – 1)/2 = t From equation 1, we get x – 2 = {(y – 1)/2}² ⇒ x – 2 = (y – 1)²/4 ⇒ (y – 1)² = 4(x – 2) This represents the equation of a parabola.

Question 13. The equation of a hyperbola with foci on the x-axis is (a) x²/a² + y²/b² = 1 (b) x²/a² – y²/b² = 1 (c) x² + y² = (a² + b²) (d) x² – y² = (a² + b²)

Answer: (b) x²/a² – y²/b² = 1 Hint: The equation of a hyperbola with foci on the x-axis is defined as x²/a² – y²/b² = 1

Question 14. The equation of parabola with vertex (-2, 1) and focus (-2, 4) is (a) 10y = x² + 4x + 16 (b) 12y = x² + 4x + 16 (c) 12y = x² + 4x (d) 12y = x² + 4x + 8

Answer: (b) 12y = x² + 4x + 16 Hint: Given, parabola having vertex is (-2, 1) and focus is (-2, 4) As the vertex and focus share the same abscissa i.e. -2, parabola axis of symmetry as x = -2 ⇒ x + 2 = 0 Hence, the equation of a parabola is of the type (y – k) = a(x – h)² where (h, k) is vertex Now, focus = (h, k + 1/4a) Since, vertex is (-2, 1) and parabola passes through vertex So, focus = (-2, 1 + 1/4a) Now, 1 + 1/4a = 4 ⇒ 1/4a = 4 -1 ⇒ 1/4a = 3 ⇒ 4a = 1/3 ⇒ a = /1(3 × 4) ⇒ a = 1/12 Now, equation of parabola is (y – 1) = (1/12) × (x + 2)² ⇒ 12(y – 1) = (x + 2)² ⇒ 12y – 12 = x² + 4x + 4 ⇒ 12y = x² + 4x + 4 + 12 ⇒ 12y = x² + 4x + 16 This is the required equation of parabola.

Question 15. If a parabolic reflector is 20 cm in diameter and 5 cm deep then the focus of parabolic reflector is (a) (0 0) (b) (0, 5) (c) (5, 0) (d) (5, 5)

Question 16. The radius of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is? (a) √57/4 (b) √77/4 (c) √77/2 (d) √87/4

Answer: (c) √77/2 Hint: Given, equation fo the of the circle is 4x² + 4y² – 8x + 12y – 25 = 0 ⇒ x² + y² – 8x/4 + 12y/4 – 25/4 = 0 ⇒ x² + y² – 2x + 3y – 25/4 = 0 Now, radius = √{(-2)² + (3)² – (-25/4)} = √{4 + 9 + 25/4} = √{13 + 25/4} = √{(13×4 + 25)/4} = √{(52 + 25)/4} = √{77/4} = √77/2

Question 17. If (a, b) is the mid point of a chord passing through the vertex of the parabola y² = 4x, then (a) a = 2b (b) 2a = b (c) a² = 2b (d) 2a = b²

Answer: (d) 2a = b² Hint: Let P(x, y) be the coordinate of the other end of the chord OP where O(0, 0) Now, (x + 0)/2 = a ⇒ x = 2a and (y + 0)/2 = b ⇒ y = 2b Now, y² = 4x ⇒ (2b)² = 4 × 2a ⇒ 4b² = 8a ⇒ b² = 2a

Question 18. A rod of length 12 CM moves with its and always touching the co-ordinate Axes. Then the equation of the locus of a point P on the road which is 3 cm from the end in contact with the x-axis is (a) x²/81 + y²/9 = 1 (b) x²/9 + y²/81 = 1 (c) x²/169 + y²/9 = 1 (d) x²/9 + y²/169 = 1

Question 19. The line lx + my + n = 0 will touches the parabola y² = 4ax if (a) ln = am² (b) ln = am (c) ln = a² m² (d) ln = a² m

Answer: (a) ln = am² Hint: Given, lx + my + n = 0 ⇒ my = -lx – n ⇒ y = (-l/m)x + (-n/m) This will touches the parabola y² = 4ax if (-n/m) = a/(-l/m) ⇒ (-n/m) = (-am/l) ⇒ n/m = am/l ⇒ ln = am²

Question 20. The center of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is? (a) (2,-3) (b) (-2,3) (c) (-4,6) (d) (4,-6)

Answer: (a) (2,-3) Hint: Given, equation fo the of the circle is 4x² + 4y² – 8x + 12y – 25 = 0 ⇒ x² + y² – 8x/4 + 12y/4 – 25/4 = 0 ⇒ x² + y² – 2x + 3y – 25/4 = 0 Now, center = {-(-2), -3} = (2, -3)

NCERT Solutions Class 11 Maths Chapter 11 Conic Sections

NCERT solutions for Class 11 maths Chapter 11 Conic Sections sheds light on the types of curves commonly known as ellipses, parabolas, circles, and hyperbolas. These curves are known as conic sections or conics, which are obtained when a plane intersects with a double-napped right circular cone . Such curves are important in a wide range of applications, such as in the reflectors of flashlights, the study of planetary motion, antenna and telescope design. These NCERT solutions Class 11 maths Chapter 11 aim at providing information about the different types of curves as mentioned above with relevant practical examples.

Various kinds of conic sections are obtained depending on where the intersecting plane with respect to the cone is and what angle it makes with the vertical axis of the cone . This intersection can be at the vertex of the cone or on any other part of the nappe, above or below the vertex . These curves are critical tools for designing applications to explore space as well as study the behavior of atomic particles. More facts and information about these can be read in the pdf file of the Class 11 maths NCERT solutions Chapter 11 Conic Sections given below and also download the exercise-wise solutions provided in the links below.

• NCERT Solutions Class 11 Maths Chapter 11 Ex 11.1
• NCERT Solutions Class 11 Maths Chapter 11 Ex 11.2
• NCERT Solutions Class 11 Maths Chapter 11 Ex 11.3
• NCERT Solutions Class 11 Maths Chapter 11 Ex 11.4
• NCERT Solutions Class 11 Maths Chapter 11 Miscellaneous Ex

NCERT Solutions for Class 11 Maths Chapter 11 PDF 

The knowledge of various kinds of curves will help the students not only in mathematics but also in the understanding of other subjects like physics; hence it is advisable for the kids to follow up on the basic facts and formulas as closely as possible. The pdf for the different exercises in the NCERT Solutions Class 11 Maths are given below :

☛ Download Class 11 Maths NCERT Solutions Chapter 11 Conic Sections

NCERT Class 11 Maths Chapter 11   Download PDF

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

This chapter aims at exploring the various conic sections and explaining them with the help of well-illustrated diagrams. In-depth knowledge of the different curves has been provided in a lucid manner throughout, which the students must make use of by reading it carefully. This chapter consists of a number of formulas applied to different kinds of questions hence, children must make it a point to periodically revise them. An exercise-wise detailed analysis of NCERT Solutions Class 11 Maths Chapter 11 Conic Sections is given below :

• Class 11 Maths Chapter 11 Ex 11.1 - 15 Questions
• Class 11 Maths Chapter 11 Ex 11.2 - 12 Questions
• Class 11 Maths Chapter 11 Ex 11.3 - 20 Questions
• Class 11 Maths Chapter 11 Ex 11.4 - 15 Questions
• Class 11 Maths Chapter 11 Miscellaneous Ex 11.4 - 8 Questions

☛  Download Class 11 Maths Chapter 11 NCERT Book

Topics Covered: The Class 11 maths NCERT solutions Chapter 11 covers the topics like circle , ellipse , parabola , hyperbola , degenerated conic sections , standard equations of such curves, the relationship between semi-major axis, semi-minor axis, and the distance of the focus from the center of the ellipse, as well as special cases of the ellipse. Further topics like eccentricity, and latus rectum are also explained.

Total Questions: Class 11 maths Chapter 11 Conic Sections consists of a total 70 questions of which 55 are easy, 10 are moderate and the remaining are long answer type questions.

List of Formulas in NCERT Solutions Class 11 Maths Chapter 11

The different curves talked about in this chapter have their own unique standard equations, as well as many formulas to determine the different parameters. Students are advised to pay attention to the derivation of these formulas as well, which will benefit them in understanding the underlying concepts. It is highly recommended that kids make well-organized formula charts which will give them a quick recap of these concepts whenever required. The key to solving this chapter successfully is by memorizing the formulas . Some important ones covered in NCERT solutions for Class 11 maths Chapter 11 are given below :

• Standard Equation of Circle : If C (h, k) is the center and r is the radius of a circle, while P(x, y) is any point on the circle, then (x – h) 2 + (y – k) 2 = r 2
• Standard Equation of Ellipse : x 2 /b 2 + y 2 /a 2 = 1 wherein the center of the ellipse is the origin and ‘b’ and ‘a’ are the X and Y intercepts respectively.
• Equation of hyperbola with foci on x-axis : x 2 /a 2 - y 2 /b 2 = 1 where center is at the origin and ‘a’ is the distance from center to vertex and ‘c’ is the distance from center to foci and b 2 = c 2 - a 2

FAQs on NCERT Solutions Class 11 Maths Chapter 11

Why are class 11 maths ncert solutions chapter 11 important.

The NCERT aims at providing easy access to knowledge for everyone through simple language. Hence, NCERT Solutions Class 11 Maths Chapter 11 are well researched by eminent scholars to present information in a relatable manner to the students with the help of solved practical examples. Also, the CBSE board highly recommends NCERT books for reference, making them an important learning resource.

Do I Need to Practice all Questions Provided in NCERT Solutions Class 11 Maths Conic Sections?

Mathematics, as we know, always gets better with practice. Regular practice helps with the consistent revision of concepts through trial and error. With the ample number of solved examples and questions provided in the NCERT Solutions Class 11 Maths Conic Sections, it is a good idea that the students make it a habit to solve all of these. This will increase their problem-solving skills giving them enough confidence to deal with any sort of questions related to curves.

What are the Important Topics Covered in NCERT Solutions Class 11 Maths Chapter 11?

The NCERT Solutions Class 11 Maths Chapter 11 explains the formation of different kinds of curves like, circle, ellipse, parabola, and hyperbola. It also talks about the standard equation of all of these curves. Further, it introduces the terms related to these curves, like the latus rectum, and eccentricity. Kids need to first understand the terms related to these curves and then move on to learning the formulas so as to score the best possible marks.

How Many Questions are there in NCERT Solutions Class 11 Maths Chapter 11 Conic Sections?

There are in total 70 questions in the NCERT Solutions Class 11 Maths Chapter 11 Conic Sections which have 55 easy problems, 10 moderately easy while the remaining are a bit difficult. The students must make note of all the important formulas given throughout the book to ensure that they can progress through these sums smoothly.

Why Should I Practice Class 11 Maths NCERT Solutions Conic Sections Chapter 11?

The NCERT Solutions Class 11 Maths Conic Sections Chapter 10 will help students improve their problem-solving and mental math skills. These answers will provide them with enough practice on working with the equations of different types of curves, be it parabola, circle or ellipse. These topics form a major foundation of the topics taught in the higher classes, thus it is critical to building their fundamentals from this chapter. Hence, kids must practice each and every question.

How CBSE Students can utilize NCERT Solutions Class 11 Maths Chapter 11 effectively?

The NCERT Solutions Class 11 Maths Chapter 11 has well-researched content on the different types of curves written by notable scholars in simple language. This makes it easy for the students to understand not only the logic behind these curves but also their practical implications. There are extensive solved examples and questions which the students must practice every day. Also, the important formulas and facts are offered in the chapter's highlights section at the end, which students can refer to when in doubt. In this way, they can utilize this resource effectively.

Conic Sections Class 11th: Notes, Mind Map and Extra Q&A

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Welcome to our comprehensive guide on Conic Sections for Class 11th students! In this article, we will provide you with detailed class notes, mind maps, and extra questions and answers to help you master this important topic in mathematics. Conic sections are a fundamental part of geometry, and understanding them is crucial for solving problems related to curves and their properties.

Whether you are studying for your exams or simply want to deepen your knowledge, this article is here to help. Our class notes cover all the important concepts, including the definitions of conic sections, their equations, properties, and practical applications. Additionally, we have included mind maps that visually summarize the key points, making it easier for you to organize and remember the information.

To ensure that you have a thorough understanding, we have also compiled a set of extra questions and answers. These practice questions will test your grasp of the topic and help you identify any areas that require further attention. With our comprehensive resources, you'll be well-prepared to tackle conic sections with confidence. Let's delve into the world of curves and shapes together!

Class 11 Maths Chapter 10 introduces students to the captivating world of conic sections, an essential topic in geometry and algebra. This chapter is a vital part of the curriculum for Class 11 students, offering an in-depth exploration of the curves obtained by intersecting a plane with a cone. Understanding conic sections in Class 11th is not just about learning geometrical shapes; it's about appreciating the underlying mathematical principles that describe various natural and man-made phenomena.

In Class 11, conic sections cover the study of parabolas, ellipses, and hyperbolas. These are the curves formed when a right circular cone is cut by a plane at different angles. The conic sections class 11th chapter is designed to give students a comprehensive understanding of these shapes, their properties, and their equations. This understanding is crucial for solving complex problems in geometry and is also applicable in various fields like physics, engineering, and astronomy.

The class 11th conic section miscellaneous exercise at the end of the chapter provides students with a variety of problems to practice and master the concepts. These exercises challenge students to apply their understanding in diverse scenarios, enhancing their problem-solving skills.

For a thorough understanding of the topic, conic sections class 11 all formulas are provided. These formulas are the backbone of the chapter, encompassing the standard equations and properties of parabolas, ellipses, and hyperbolas. Students can access conic sections class 11 notes in PDF format for a detailed explanation of these concepts. These notes, often available for download, are an excellent resource for in-depth study and revision.

Additionally, conic sections class 11 PDF notes are a great way for students to study on-the-go. They include key points, illustrations, and solved examples, making the learning process more engaging and effective.

For visual learners, a conic sections class 11 mind map can be extremely helpful. It visually organizes and connects various concepts within the chapter, making revision more efficient and effective. Moreover, conic sections class 11 MCQs (Multiple Choice Questions) are an effective tool for self-assessment. These MCQs cover various aspects of the chapter, ensuring a comprehensive review and helping students prepare for their exams.

In summary, Chapter 10 of Class 11 Maths is crucial for students aiming to excel in mathematics. It not only enhances their geometrical understanding but also prepares them for advanced studies and applications in various fields. The combination of detailed notes, mind maps, formula lists, miscellaneous exercises, and MCQs ensures that students have a holistic understanding of conic sections, setting a strong foundation for their future academic endeavors.

• Conic sections class 11th:
• All CBSE notes
• All MATHS notes
• All 11 SCIENCE notes

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

LEVEL UP CBSE Question Bank for Class 11 Maths Chapter 11 Conic Sections  will provide you with  detailed, latest, comprehensive  &  confidence inspiring   solutions  to the maximum number of Questions covering all the  topics from  your  NCERT Text Books !

Given below are the  Important Questions for Class 11 Maths  (with Solutions) from the exam point of view. To download the “ CBSE Question Bank for Class 11 Maths”,  just click on the “ Download PDF ” and “ Download Solutions ” buttons

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

Checkout our cbse question banks for other chapters.

• Chapter 9 Sequences and Series CBSE Question Bank
• Chapter 10 Straight Lines CBSE Question Bank
• Chapter 12 Introduction to 3D Geometry CBSE Question Bank
• Chapter 13 Limits and Derivatives CBSE Question Bank

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections

Ncert solutions for class 11 maths chapter 11 – conic sections pdf.

Free PDF of NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 11 – Conic Sections Maths NCERT Solutions for Class 11 to help you to score more marks in your board exams and as well as competitive exams.

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• NCERT Solutions
• NCERT Class 11
• NCERT 11 Maths
• Chapter 11: Conic Sections
• Chapter 11 Conic Sections Miscellaneous Ex

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Miscellaneous Exercise

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

NCERT Solutions for Class 11 Maths are provided here to help the students in understanding the steps to solve mathematical problems that are in the Class 11 Maths NCERT textbook. Chapter 11 Conic Sections of Class 11 Maths is categorised under the CBSE Syllabus for the session 2023-24. The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections is based on the following topics:

• Sections of a Cone

Each question of the exercise has been carefully solved for the students to understand, keeping the examination in mind. The NCERT Solutions for Class 11 Maths helps the students in understanding all the concepts of Class 11 Maths, in-depth.

NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections Miscellaneous Exercise

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Solutions for Class 11 Maths Chapter 11 – Miscellaneous Exercise

1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

We know that the origin of the coordinate plane is taken at the vertex of the parabolic reflector, where the axis of the reflector is along the positive x – axis.

Diagrammatic representation is as follows:

We know that the equation of the parabola is of the form y 2  = 4ax (as it is opening to the right)

Since, the parabola passes through point A(10, 5),

10 2  = 4a(5)

The focus of the parabola is (a, 0) = (5, 0), which is the mid – point of the diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

We know that the origin of the coordinate plane is taken at the vertex of the arch, where its vertical axis is along the positive y –axis.

The equation of the parabola is of the form x 2  = 4ay (as it is opening upwards).

It is given that at base arch is 10 m high and 5 m wide.

So, y = 10 and x = 5/2 from the above figure.

It is clear that the parabola passes through point (5/2, 10)

So, x 2  = 4ay

(5/2) 2 = 4a(10)

4a = 25/(4×10)

we know the arch is in the form of a parabola whose equation is x 2 = 5/8y

We need to find width, when height = 2 m.

To find x, when y = 2.

When, y = 2,

x 2 = 5/8 (2)

AB = 2 × √5/2 m

= 2.23 m (approx.)

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.

3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

We know that the vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y –axis.

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadways, 18 m from the middle.

So, AB = 30 m, OC = 6m, and BC = 50 m.

The equation of the parabola is x 2 = 4ay (as it is opening upwards).

The coordinates of point A are (50, 30 -6) = (50, 24)

Since, A(50, 24) is a point on the parabola.

(50) 2  = 4a(24)

a = (50×50)/(4×24)

Equation of the parabola, x 2 = 4ay = 4×(625/24)y or 6x 2 = 625y

The x – coordinate of point D is 18.

Hence, at x = 18,

6(18) 2  = 625y

y = (6×18×18)/625

= 3.11 (approx.)

Thus, DE = 3.11 m

DF = DE +EF = 3.11 m +6 m = 9.11 m

Hence, the length of the supporting wire attached to the roadway 18 m from the middle is approximately 9.11 m.

4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Since, the height and width of the arc from the centre is 2 m and 8 m, respectively, it is clear that the length of the major axis is 8 m, while the length of the semi- minor axis is 2 m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis.

Hence, Diagrammatic representation of semi- ellipse is as follows:

The equation of the semi – ellipse will be of the from x 2 /16 + y 2 /4 = 1, y ≥ 0 … (1

Let A be a point on the major axis such that AB = 1.5m.

Now draw AC ⊥ OB.

OA = (4 – 1.5)m = 2.5m

The x – coordinate of point C is 2.5

On substituting the value of x with 2.5 in equation (1), we get,

(2.5) 2 /16 + y 2 /4 = 1

6.25/16 + y 2 /4 = 1

y 2 = 4 (1 – 6.25/16)

= 4 (9.75/16)

y = 1.56 (approx.)

So, AC = 1.56m

Hence, the height of the arch at a point 1.5m from one end is approximately 1.56m.

5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Let AB be the rod making an angle Ɵ with OX and P(x,y) be the point on it such that

Then, PB = AB – AP = (12 – 3) cm = 9cm [AB = 12cm]

From P, draw PQ ⊥ OY and PR ⊥ OX.

In ΔPBQ, cos θ = PQ/PB = x/9

Sin θ = PR/PA = y/3

we know that, sin 2 θ +cos 2 θ = 1,

(y/3) 2 + (x/9) 2 = 1 or

x 2 /81 + y 2 /9 = 1

Hence, the equation of the locus of point P on the rod is x 2 /81 + y 2 /9 = 1

6. Find the area of the triangle formed by the lines joining the vertex of the parabola x 2  = 12y to the ends of its latus rectum.

The given parabola is x 2  = 12y.

On comparing this equation with x 2 = 4ay, we get,

The coordinates of foci are S(0,a) = S(0,3).

Now let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At y = 3, x 2  = 12(3)

So, the coordinates of A are (-6, 3), while the coordinates of B are (6, 3)

Then, the vertices of ΔOAB are O(0,0), A (-6,3) and B(6,3).

By using the formula,

Area of ΔOAB = ½ [0(3-3) + (-6)(3-0) + 6(0-3)] unit 2

= ½ [(-6) (3) + 6 (-3)] unit 2

= ½ [-18-18] unit 2

= ½ [-36] unit 2

= 18 unit 2

∴ Area of ΔOAB is 18 unit 2

7. A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man.

So, PA + PB = 10.

We know that if a point moves in plane in such a way that the sum of its distance from two fixed point is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Then, the path described by the man is an ellipse where the length of the major axis is 10m, while points A and B are the foci.

Now let us take the origin of the coordinate plane as the centre of the ellipse, and taking the major axis along the x- axis,

The diagrammatic representation of the ellipse is as follows:

The equation of the ellipse is in the form of x 2 /a 2 + y 2 /b 2 = 1, where ‘a’ is the semi-major axis.

So, 2a = 10

Distance between the foci, 2c = 8

By using the relation, c = √ (a 2 – b 2 ), we get,

4 = √ (25 – b 2 )

16 = 25 – b 2

b 2  = 25 -1

Hence, equation of the path traced by the man is x 2 /25 + y 2 /9 = 1

8. An equilateral triangle is inscribed in the parabola y 2  = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Let us consider OAB be the equilateral triangle inscribed in parabola y 2  = 4ax.

Let AB intersect the x – axis at point C.

Diagrammatic representation of the ellipse is as follows:

Now let OC = k

From the equation of the given parabola, we have,

So, y 2  = 4ak

y = ± 2 √ ak

The coordinates of points A and B are (k, 2 √ ak), and (k, -2 √ ak)

AB = CA + CB

= 2 √ ak + 2 √ ak

Since, OAB is an equilateral triangle, OA 2  = AB 2 .

k 2 + (2 √ ak) 2 = (4 √ ak) 2

k 2  + 4ak = 16ak

k 2  = 12ak

Thus, AB = 4 √ ak = 4 √ (a×12a)

= 4 √ 12a 2

= 4 √ (4a×3a)

= 4(2) √ 3a

Hence, the side of the equilateral triangle inscribed in parabola y 2  = 4ax is 8 √ 3a.

Access Other Exercise Solutions of Class 11 Maths Chapter 11 – Conic Sections

Exercise 11.1 Solutions 15 Questions

Exercise 11.2 Solutions 12 Questions

Exercise 11.3 Solutions 20 Questions

Exercise 11.4 Solutions 15 Questions

Also explore – NCERT Class 11 Solutions

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