NCERT Solutions for Class 11 Maths Chapter 10: Conic Sections

These Class 11 Maths Chapter 10 solutions cover Conic Sections — circles, parabolas, ellipses and hyperbolas. Every question of Exercise 10.1, 10.2, 10.3, 10.4 and the Miscellaneous Exercise is reproduced exactly as in the NCERT textbook and solved step by step, with foci, vertices, eccentricity and latus rectum worked out and verified against the book’s answer key (Session 2026–27).

Class: 11 Subject: Mathematics Chapter: 10 – Conic Sections Exercises: 10.1, 10.2, 10.3, 10.4 + Miscellaneous Session: 2026–27
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Chapter 10 Overview

Chapter 10 of Class 11 Maths, Conic Sections, studies the curves obtained when a plane cuts a double-napped right circular cone: the circle, parabola, ellipse and hyperbola. Each conic is defined by a geometric property (equidistance, sum of distances, difference of distances) and is reduced to a neat standard equation when its centre or vertex is placed at the origin. You learn to write the equation of a circle from its centre and radius, to find the focus, axis, directrix and latus rectum of a parabola, and to compute the foci, vertices, axes lengths, eccentricity and latus rectum of ellipses and hyperbolas. The chapter ends with real-life modelling problems — parabolic reflectors, suspension-bridge cables and semi-elliptical arches. The Class 11 Maths Chapter 10 solutions below solve every exercise question and the Miscellaneous Exercise in full.

Key Concepts & Definitions

Circle: the set of all points in a plane that are equidistant (radius r) from a fixed point (the centre).

Parabola: the set of all points equidistant from a fixed line (the directrix) and a fixed point not on it (the focus). The axis passes through the focus perpendicular to the directrix; the vertex is where the parabola meets its axis.

Ellipse: the set of all points the sum of whose distances from two fixed points (the foci) is a constant 2a. It has a major axis (length 2a), a minor axis (length 2b) and a centre at the mid-point of the foci.

Hyperbola: the set of all points the difference of whose distances from two fixed points (the foci) is a constant 2a. It has a transverse axis (length 2a) and a conjugate axis (length 2b).

Eccentricity (e): e = c/a, the ratio of the centre-to-focus distance to the centre-to-vertex distance. For an ellipse 0 < e < 1; for a hyperbola e > 1.

Latus rectum: the chord through a focus perpendicular to the major/transverse axis, with both ends on the conic.

Important Formulas (Chapter 10)

Circle: centre (h, k), radius r → (x − h)2 + (y − k)2 = r2. General form x2 + y2 + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g2 + f2 − c).

Parabola (standard): y2 = 4ax (opens right), y2 = −4ax (left), x2 = 4ay (up), x2 = −4ay (down). For y2 = 4ax: focus (a, 0), directrix x = −a, axis the x-axis, latus rectum = 4a.

Ellipse: x2/a2 + y2/b2 = 1 (a > b, major axis on x-axis), with c2 = a2 − b2, foci (±c, 0), vertices (±a, 0), e = c/a, latus rectum = 2b2/a.

Hyperbola: x2/a2 − y2/b2 = 1, with c2 = a2 + b2, foci (±c, 0), vertices (±a, 0), e = c/a (≥ 1), latus rectum = 2b2/a.

Exercise 10.1 Solutions (Circle)

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the book’s Answers appendix.

In each of the following Exercises 1 to 5, find the equation of the circle with

1. centre (0, 2) and radius 2

SOLUTION (x − h)2 + (y − k)2 = r2 with h = 0, k = 2, r = 2. (x − 0)2 + (y − 2)2 = 22 ⇒ x2 + y2 − 4y + 4 = 4. ∴ x2 + y2 − 4y = 0.

2. centre (−2, 3) and radius 4

SOLUTION (x + 2)2 + (y − 3)2 = 42 = 16. x2 + 4x + 4 + y2 − 6y + 9 = 16. ∴ x2 + y2 + 4x − 6y − 3 = 0.

3. centre (½, ¼) and radius 1/12

SOLUTION (x − ½)2 + (y − ¼)2 = (1/12)2 = 1/144. x2 − x + ¼ + y2 − ½y + 1/16 = 1/144. Multiply through by 144: 144x2 + 144y2 − 144x − 72y + 36 + 9 = 1. ∴ 36x2 + 36y2 − 36x − 18y + 11 = 0 (dividing by 4).

4. centre (1, 1) and radius √2

SOLUTION (x − 1)2 + (y − 1)2 = (√2)2 = 2. x2 − 2x + 1 + y2 − 2y + 1 = 2. ∴ x2 + y2 − 2x − 2y = 0.

5. centre (−a, −b) and radius √(a2 − b2)

SOLUTION (x + a)2 + (y + b)2 = a2 − b2. x2 + 2ax + a2 + y2 + 2by + b2 = a2 − b2. ∴ x2 + y2 + 2ax + 2by + 2b2 = 0.

In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)2 + (y − 3)2 = 36

SOLUTION Compare with (x − h)2 + (y − k)2 = r2: h = −5, k = 3, r2 = 36. ∴ centre (−5, 3), radius 6.

7. x2 + y2 − 4x − 8y − 45 = 0

SOLUTION Group and complete squares: (x2 − 4x) + (y2 − 8y) = 45. (x − 2)2 − 4 + (y − 4)2 − 16 = 45 ⇒ (x − 2)2 + (y − 4)2 = 65. ∴ centre (2, 4), radius √65.

8. x2 + y2 − 8x + 10y − 12 = 0

SOLUTION (x2 − 8x) + (y2 + 10y) = 12. (x − 4)2 − 16 + (y + 5)2 − 25 = 12 ⇒ (x − 4)2 + (y + 5)2 = 53. ∴ centre (4, −5), radius √53.

9. 2x2 + 2y2 − x = 0

SOLUTION Divide by 2: x2 + y2 − ½x = 0 ⇒ (x2 − ½x) + y2 = 0. (x − ¼)2 − 1/16 + y2 = 0 ⇒ (x − ¼)2 + y2 = 1/16. ∴ centre (¼, 0), radius ¼.

10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

SOLUTION Let the centre be (h, k). Equal distances from (4, 1) and (6, 5): (h − 4)2 + (k − 1)2 = (h − 6)2 + (k − 5)2. Expand: −8h + 16 − 2k + 1 = −12h + 36 − 10k + 25 ⇒ 4h + 8k = 44 ⇒ h + 2k = 11.  …(i) Centre on the line: 4h + k = 16.  …(ii) From (ii) k = 16 − 4h; sub in (i): h + 2(16 − 4h) = 11 ⇒ h + 32 − 8h = 11 ⇒ −7h = −21 ⇒ h = 3, k = 4. r2 = (3 − 4)2 + (4 − 1)2 = 1 + 9 = 10. (x − 3)2 + (y − 4)2 = 10 ⇒ x2 + y2 − 6x − 8y + 15 = 0.

11. Find the equation of the circle passing through the points (2, 3) and (−1, 1) and whose centre is on the line x − 3y − 11 = 0.

SOLUTION Let centre (h, k). Equal distances: (h − 2)2 + (k − 3)2 = (h + 1)2 + (k − 1)2. Expand: −4h + 4 − 6k + 9 = 2h + 1 − 2k + 1 ⇒ −6h − 4k + 11 = 0 ⇒ 6h + 4k = 11.  …(i) Centre on line: h − 3y − 11 = 0 ⇒ h = 3k + 11.  …(ii) Sub in (i): 6(3k + 11) + 4k = 11 ⇒ 18k + 66 + 4k = 11 ⇒ 22k = −55 ⇒ k = −5/2, h = 3(−5/2) + 11 = 7/2. r2 = (7/2 − 2)2 + (−5/2 − 3)2 = (3/2)2 + (−11/2)2 = 9/4 + 121/4 = 130/4 = 65/2. (x − 7/2)2 + (y + 5/2)2 = 65/2 ⇒ x2 + y2 − 7x + 5y − 14 = 0.

12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

SOLUTION Centre on x-axis ⇒ (h, 0). Distance to (2, 3) is 5: (h − 2)2 + (0 − 3)2 = 25. (h − 2)2 + 9 = 25 ⇒ (h − 2)2 = 16 ⇒ h − 2 = ±4 ⇒ h = 6 or h = −2. If h = 6: (x − 6)2 + y2 = 25 ⇒ x2 + y2 − 12x + 11 = 0. If h = −2: (x + 2)2 + y2 = 25 ⇒ x2 + y2 + 4x − 21 = 0.

13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

SOLUTION A circle through the origin cutting intercepts a and b passes through (0, 0), (a, 0) and (0, b). Take the general form x2 + y2 + 2gx + 2fy + c = 0. Through (0, 0): c = 0. Through (a, 0): a2 + 2ga = 0 ⇒ g = −a/2. Through (0, b): b2 + 2fb = 0 ⇒ f = −b/2. ∴ x2 + y2 − ax − by = 0, i.e. x2 + y2 − ax − by = 0.

14. Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

SOLUTION r2 = (4 − 2)2 + (5 − 2)2 = 4 + 9 = 13. (x − 2)2 + (y − 2)2 = 13 ⇒ x2 − 4x + 4 + y2 − 4y + 4 = 13. ∴ x2 + y2 − 4x − 4y − 5 = 0 (i.e. x2 + y2 − 4x − 4y = 5).

15. Does the point (−2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

SOLUTION The circle has centre (0, 0) and radius 5. Distance of the point from the centre = √((−2.5)2 + (3.5)2) = √(6.25 + 12.25) = √18.5 ≈ 4.30. Since 4.30 < 5, the distance is less than the radius. ∴ the point lies inside the circle.

Exercise 10.2 Solutions (Parabola)

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

1. y2 = 12x

SOLUTION Compare with y2 = 4ax: 4a = 12 ⇒ a = 3. Opens right (coefficient positive). Focus (3, 0); axis the x-axis; directrix x = −3; latus rectum = 4a = 12.

2. x2 = 6y

SOLUTION Compare with x2 = 4ay: 4a = 6 ⇒ a = 3/2. Opens upward. Focus (0, 3/2); axis the y-axis; directrix y = −3/2; latus rectum = 4a = 6.

3. y2 = −8x

SOLUTION Compare with y2 = −4ax: 4a = 8 ⇒ a = 2. Opens left. Focus (−2, 0); axis the x-axis; directrix x = 2; latus rectum = 4a = 8.

4. x2 = −16y

SOLUTION Compare with x2 = −4ay: 4a = 16 ⇒ a = 4. Opens downward. Focus (0, −4); axis the y-axis; directrix y = 4; latus rectum = 4a = 16.

5. y2 = 10x

SOLUTION Compare with y2 = 4ax: 4a = 10 ⇒ a = 5/2. Opens right. Focus (5/2, 0); axis the x-axis; directrix x = −5/2; latus rectum = 4a = 10.

6. x2 = −9y

SOLUTION Compare with x2 = −4ay: 4a = 9 ⇒ a = 9/4. Opens downward. Focus (0, −9/4); axis the y-axis; directrix y = 9/4; latus rectum = 4a = 9.

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

7. Focus (6, 0); directrix x = −6

SOLUTION Focus on positive x-axis and directrix x = −6 ⇒ parabola y2 = 4ax with a = 6. ∴ y2 = 24x.

8. Focus (0, −3); directrix y = 3

SOLUTION Focus on negative y-axis and directrix y = 3 ⇒ parabola x2 = −4ay with a = 3. ∴ x2 = −12y.

9. Vertex (0, 0); focus (3, 0)

SOLUTION Focus on positive x-axis ⇒ y2 = 4ax with a = 3. ∴ y2 = 12x.

10. Vertex (0, 0); focus (−2, 0)

SOLUTION Focus on negative x-axis ⇒ y2 = −4ax with a = 2. ∴ y2 = −8x.

11. Vertex (0, 0) passing through (2, 3) and axis is along x-axis.

SOLUTION Axis along x-axis, vertex at origin, point in first quadrant ⇒ y2 = 4ax. Pass through (2, 3): 32 = 4a(2) ⇒ 9 = 8a ⇒ 4a = 9/2. ∴ y2 = (9/2)x, i.e. 2y2 = 9x.

12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.

SOLUTION Symmetric about y-axis, opens upward (point has positive y) ⇒ x2 = 4ay. Pass through (5, 2): 52 = 4a(2) ⇒ 25 = 8a ⇒ 4a = 25/2. ∴ x2 = (25/2)y, i.e. 2x2 = 25y.

Exercise 10.3 Solutions (Ellipse)

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

1. x2/36 + y2/16 = 1

SOLUTION a2 = 36, b2 = 16 (major axis on x-axis). a = 6, b = 4, c = √(36 − 16) = √20. Foci (±√20, 0); vertices (±6, 0); major axis = 12; minor axis = 8; e = c/a = √20/6; latus rectum = 2b2/a = 32/6 = 16/3.

2. x2/4 + y2/25 = 1

SOLUTION Larger denominator under y2 ⇒ major axis on y-axis. a2 = 25, b2 = 4, a = 5, b = 2, c = √(25 − 4) = √21. Foci (0, ±√21); vertices (0, ±5); major axis = 10; minor axis = 4; e = √21/5; latus rectum = 2b2/a = 8/5 = 8/5.

3. x2/16 + y2/9 = 1

SOLUTION a2 = 16, b2 = 9 (major on x-axis). a = 4, b = 3, c = √(16 − 9) = √7. Foci (±√7, 0); vertices (±4, 0); major axis = 8; minor axis = 6; e = √7/4; latus rectum = 2b2/a = 18/4 = 9/2.

4. x2/25 + y2/100 = 1

SOLUTION Larger denominator under y2 ⇒ major axis on y-axis. a2 = 100, b2 = 25, a = 10, b = 5, c = √(100 − 25) = √75. Foci (0, ±√75); vertices (0, ±10); major axis = 20; minor axis = 10; e = √75/10 = (5√3)/10 = √3/2; latus rectum = 2b2/a = 50/10 = 5.

5. x2/49 + y2/36 = 1

SOLUTION a2 = 49, b2 = 36 (major on x-axis). a = 7, b = 6, c = √(49 − 36) = √13. Foci (±√13, 0); vertices (±7, 0); major axis = 14; minor axis = 12; e = √13/7; latus rectum = 2b2/a = 72/7 = 72/7.

6. x2/100 + y2/400 = 1

SOLUTION Larger denominator under y2 ⇒ major axis on y-axis. a2 = 400, b2 = 100, a = 20, b = 10, c = √(400 − 100) = √300 = 10√3. Foci (0, ±10√3); vertices (0, ±20); major axis = 40; minor axis = 20; e = c/a = 10√3/20 = √3/2; latus rectum = 2b2/a = 200/20 = 10.

7. 36x2 + 4y2 = 144

SOLUTION Divide by 144: x2/4 + y2/36 = 1. Major axis on y-axis. a2 = 36, b2 = 4, a = 6, b = 2, c = √(36 − 4) = √32 = 4√2. Foci (0, ±4√2); vertices (0, ±6); major axis = 12; minor axis = 4; e = c/a = 4√2/6 = 2√2/3; latus rectum = 2b2/a = 8/6 = 4/3.

8. 16x2 + y2 = 16

SOLUTION Divide by 16: x2/1 + y2/16 = 1. Major axis on y-axis. a2 = 16, b2 = 1, a = 4, b = 1, c = √(16 − 1) = √15. Foci (0, ±√15); vertices (0, ±4); major axis = 8; minor axis = 2; e = √15/4; latus rectum = 2b2/a = 2/4 = 1/2.

9. 4x2 + 9y2 = 36

SOLUTION Divide by 36: x2/9 + y2/4 = 1. Major axis on x-axis. a2 = 9, b2 = 4, a = 3, b = 2, c = √(9 − 4) = √5. Foci (±√5, 0); vertices (±3, 0); major axis = 6; minor axis = 4; e = √5/3; latus rectum = 2b2/a = 8/3 = 8/3.

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

10. Vertices (±5, 0), foci (±4, 0)

SOLUTION Vertices on x-axis ⇒ x2/a2 + y2/b2 = 1, a = 5, c = 4. b2 = a2 − c2 = 25 − 16 = 9. ∴ x2/25 + y2/9 = 1.

11. Vertices (0, ±13), foci (0, ±5)

SOLUTION Vertices on y-axis ⇒ x2/b2 + y2/a2 = 1, a = 13, c = 5. b2 = 169 − 25 = 144. ∴ x2/144 + y2/169 = 1.

12. Vertices (±6, 0), foci (±4, 0)

SOLUTION a = 6, c = 4 (x-axis). b2 = 36 − 16 = 20. ∴ x2/36 + y2/20 = 1.

13. Ends of major axis (±3, 0), ends of minor axis (0, ±2)

SOLUTION Major axis on x-axis ⇒ a = 3, b = 2. ∴ x2/9 + y2/4 = 1.

14. Ends of major axis (0, ±√5), ends of minor axis (±1, 0)

SOLUTION Major axis on y-axis ⇒ a = √5 (a2 = 5), b = 1 (b2 = 1). ∴ x2/1 + y2/5 = 1.

15. Length of major axis 26, foci (±5, 0)

SOLUTION 2a = 26 ⇒ a = 13; foci on x-axis ⇒ c = 5. b2 = 169 − 25 = 144. ∴ x2/169 + y2/144 = 1.

16. Length of minor axis 16, foci (0, ±6).

SOLUTION 2b = 16 ⇒ b = 8 (b2 = 64); foci on y-axis ⇒ c = 6. a2 = b2 + c2 = 64 + 36 = 100. ∴ x2/64 + y2/100 = 1.

17. Foci (±3, 0), a = 4

SOLUTION Foci on x-axis ⇒ c = 3, a = 4. b2 = a2 − c2 = 16 − 9 = 7. ∴ x2/16 + y2/7 = 1.

18. b = 3, c = 4, centre at the origin; foci on the x axis.

SOLUTION Foci on x-axis ⇒ major axis on x-axis. a2 = b2 + c2 = 9 + 16 = 25. ∴ x2/25 + y2/9 = 1.

19. Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

SOLUTION Major axis on y-axis ⇒ x2/b2 + y2/a2 = 1. Let 1/b2 = p, 1/a2 = q. Through (3, 2): 9p + 4q = 1. Through (1, 6): p + 36q = 1. From the second, p = 1 − 36q; sub: 9(1 − 36q) + 4q = 1 ⇒ 9 − 324q + 4q = 1 ⇒ −320q = −8 ⇒ q = 1/40, p = 1 − 36/40 = 1/10. So b2 = 10, a2 = 40. ∴ x2/10 + y2/40 = 1.

20. Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

SOLUTION Major axis on x-axis ⇒ x2/a2 + y2/b2 = 1. Let 1/a2 = p, 1/b2 = q. Through (4, 3): 16p + 9q = 1. Through (6, 2): 36p + 4q = 1. Solve: multiply first by 4 (64p + 36q = 4) and second by 9 (324p + 36q = 9); subtract: 260p = 5 ⇒ p = 1/52. Then 16/52 + 9q = 1 ⇒ 9q = 1 − 4/13 = 9/13 ⇒ q = 1/13. So a2 = 52, b2 = 13. ∴ x2/52 + y2/13 = 1 (i.e. x2 + 4y2 = 52).

Exercise 10.4 Solutions (Hyperbola)

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

1. x2/16 − y2/9 = 1

SOLUTION a2 = 16, b2 = 9 (transverse axis on x-axis). a = 4, b = 3, c = √(16 + 9) = 5. Foci (±5, 0); vertices (±4, 0); e = c/a = 5/4; latus rectum = 2b2/a = 18/4 = 9/2.

2. y2/9 − x2/27 = 1

SOLUTION Positive term is y2 ⇒ transverse axis on y-axis. a2 = 9, b2 = 27, a = 3, b = 3√3, c = √(9 + 27) = 6. Foci (0, ±6); vertices (0, ±3); e = c/a = 2; latus rectum = 2b2/a = 54/3 = 18.

3. 9y2 − 4x2 = 36

SOLUTION Divide by 36: y2/4 − x2/9 = 1. Transverse axis on y-axis. a2 = 4, b2 = 9, a = 2, b = 3, c = √(4 + 9) = √13. Foci (0, ±√13); vertices (0, ±2); e = c/a = √13/2; latus rectum = 2b2/a = 18/2 = 9.

4. 16x2 − 9y2 = 576

SOLUTION Divide by 576: x2/36 − y2/64 = 1. Transverse axis on x-axis. a2 = 36, b2 = 64, a = 6, b = 8, c = √(36 + 64) = 10. Foci (±10, 0); vertices (±6, 0); e = c/a = 5/3; latus rectum = 2b2/a = 128/6 = 64/3.

5. 5y2 − 9x2 = 36

SOLUTION Divide by 36: y2/(36/5) − x2/4 = 1. Transverse axis on y-axis. a2 = 36/5, b2 = 4, a = 6/√5, c = √(36/5 + 4) = √(56/5) = 2√14/√5. Foci (0, ±2√14/√5); vertices (0, ±6/√5); e = c/a = (2√14/√5)/(6/√5) = √14/3; latus rectum = 2b2/a = 8/(6/√5) = 4√5/3.

6. 49y2 − 16x2 = 784.

SOLUTION Divide by 784: y2/16 − x2/49 = 1. Transverse axis on y-axis. a2 = 16, b2 = 49, a = 4, b = 7, c = √(16 + 49) = √65. Foci (0, ±√65); vertices (0, ±4); e = c/a = √65/4; latus rectum = 2b2/a = 98/4 = 49/2.

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

7. Vertices (±2, 0), foci (±3, 0)

SOLUTION Transverse axis on x-axis ⇒ x2/a2 − y2/b2 = 1, a = 2, c = 3. b2 = c2 − a2 = 9 − 4 = 5. ∴ x2/4 − y2/5 = 1.

8. Vertices (0, ±5), foci (0, ±8)

SOLUTION Transverse axis on y-axis ⇒ y2/a2 − x2/b2 = 1, a = 5, c = 8. b2 = 64 − 25 = 39. ∴ y2/25 − x2/39 = 1.

9. Vertices (0, ±3), foci (0, ±5)

SOLUTION Transverse axis on y-axis ⇒ y2/a2 − x2/b2 = 1, a = 3, c = 5. b2 = 25 − 9 = 16. ∴ y2/9 − x2/16 = 1.

10. Foci (±5, 0), the transverse axis is of length 8.

SOLUTION Foci on x-axis ⇒ c = 5; transverse axis 2a = 8 ⇒ a = 4. b2 = c2 − a2 = 25 − 16 = 9. ∴ x2/16 − y2/9 = 1.

11. Foci (0, ±13), the conjugate axis is of length 24.

SOLUTION Foci on y-axis ⇒ c = 13; conjugate axis 2b = 24 ⇒ b = 12 (b2 = 144). a2 = c2 − b2 = 169 − 144 = 25. ∴ y2/25 − x2/144 = 1.

12. Foci (±3√5, 0), the latus rectum is of length 8.

SOLUTION Foci on x-axis ⇒ c = 3√5 (c2 = 45); latus rectum 2b2/a = 8 ⇒ b2 = 4a. c2 = a2 + b2 ⇒ 45 = a2 + 4a ⇒ a2 + 4a − 45 = 0 ⇒ (a + 9)(a − 5) = 0 ⇒ a = 5 (positive). b2 = 4(5) = 20. ∴ x2/25 − y2/20 = 1.

13. Foci (±4, 0), the latus rectum is of length 12

SOLUTION Foci on x-axis ⇒ c = 4 (c2 = 16); 2b2/a = 12 ⇒ b2 = 6a. 16 = a2 + 6a ⇒ a2 + 6a − 16 = 0 ⇒ (a + 8)(a − 2) = 0 ⇒ a = 2. b2 = 6(2) = 12. ∴ x2/4 − y2/12 = 1.

14. vertices (±7, 0), e = 4/3.

SOLUTION Transverse axis on x-axis ⇒ a = 7. e = c/a = 4/3 ⇒ c = 28/3, c2 = 784/9. b2 = c2 − a2 = 784/9 − 49 = 784/9 − 441/9 = 343/9. ∴ x2/49 − 9y2/343 = 1 (i.e. x2/49 − y2/(343/9) = 1).

15. Foci (0, ±√10), passing through (2, 3)

SOLUTION Foci on y-axis ⇒ y2/a2 − x2/b2 = 1, with c2 = a2 + b2 = 10, so b2 = 10 − a2. Through (2, 3): 9/a2 − 4/b2 = 1 ⇒ 9/a2 − 4/(10 − a2) = 1. 9(10 − a2) − 4a2 = a2(10 − a2) ⇒ 90 − 13a2 = 10a2 − a4 ⇒ a4 − 23a2 + 90 = 0. (a2 − 5)(a2 − 18) = 0 ⇒ a2 = 5 (a2 = 18 makes b2 negative, rejected). Then b2 = 10 − 5 = 5. ∴ y2/5 − x2/5 = 1.

Miscellaneous Exercise on Chapter 10 Solutions

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

SOLUTION Place the vertex at the origin with axis along the x-axis: y2 = 4ax. At depth 5 cm, the diameter is 20 cm so the rim point is (5, 10). 102 = 4a(5) ⇒ 100 = 20a ⇒ a = 5. ∴ the focus is at (5, 0) — i.e. the focus is at the mid-point of the given diameter, 5 cm from the vertex.

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?

SOLUTION Take the vertex (top of the arch) at the origin, axis downward: x2 = −4ay (or x2 = 4Ay measuring depth y as positive downward, x2 = ky). At the base, y = 10 (below vertex) and half-width x = 2.5: (2.5)2 = k(10) ⇒ 6.25 = 10k ⇒ k = 0.625, so x2 = 0.625y. At 2 m from the vertex, y = 2: x2 = 0.625 × 2 = 1.25 ⇒ x = √1.25 ≈ 1.118. Width = 2x = 2 × 1.118 ≈ 2.23 m (approx.).

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.

SOLUTION Put the vertex (lowest point of the cable) at the origin, axis vertical: x2 = 4ay. The shortest wire (6 m) is at the centre, so above the roadway the vertex is 6 m up. At the towers, x = 50 and the cable height above the roadway is 30 m, so above the vertex it is 30 − 6 = 24 m: (50)2 = 4a(24) ⇒ 2500 = 96a ⇒ 4a = 2500/24. At x = 18: y = x2/(4a) = 182 × 24 / 2500 = 324 × 24 / 2500 = 7776/2500 = 3.11 m above the vertex. Length of the wire = 6 + 3.11 = 9.11 m (approx.).

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.

SOLUTION Centre at the origin: x2/a2 + y2/b2 = 1. Width 8 m ⇒ a = 4; height at centre 2 m ⇒ b = 2. So x2/16 + y2/4 = 1. A point 1.5 m from one end is at x = 4 − 1.5 = 2.5. y2 = 4(1 − 2.52/16) = 4(1 − 6.25/16) = 4(9.75/16) = 2.4375 ⇒ y = √2.4375 ≈ 1.56. ∴ height ≈ 1.56 m (approx.).

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.

SOLUTION Let the rod make angle θ with the x-axis, end A on the x-axis and end B on the y-axis, AB = 12. P(x, y) is 3 cm from A (the x-axis end), so PB = 9. Drop perpendiculars: from triangle with the x-axis end, cosθ = x/9; with the y-axis end, sinθ = y/3. cos2θ + sin2θ = 1 ⇒ x2/81 + y2/9 = 1. ∴ the locus is the ellipse x2/81 + y2/9 = 1.

6. Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.

SOLUTION x2 = 12y compares with x2 = 4ay: 4a = 12 ⇒ a = 3. Focus (0, 3); latus rectum is horizontal through the focus with length 4a = 12. Ends of the latus rectum: (−6, 3) and (6, 3). The vertex is (0, 0). Triangle with base 12 (between the latus-rectum ends) and height 3 (distance from vertex to the line y = 3). Area = ½ × 12 × 3 = 18 sq units.

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.

SOLUTION Constant sum of distances from two fixed points ⇒ ellipse with foci the flag posts. 2a = 10 ⇒ a = 5; distance between foci 2c = 8 ⇒ c = 4. b2 = a2 − c2 = 25 − 16 = 9. ∴ the path (taking foci on the x-axis) is x2/25 + y2/9 = 1.

8. An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

SOLUTION Let the triangle be OPQ with O the origin and P, Q symmetric about the x-axis: P = (x1, y1), Q = (x1, −y1). Side PQ = 2y1. For an equilateral triangle, OP makes 30° with the axis, so y1 = x1 tan30° = x1/√3, i.e. x1 = √3 y1. P lies on the parabola: y12 = 4a x1 = 4a(√3 y1) ⇒ y1 = 4√3 a. Side = OP = √(x12 + y12). Since x1 = √3 y1, OP = √(3y12 + y12) = 2y1 = 2(4√3 a) = 8√3 a.

Common Mistakes to Avoid

Watch out for these

  • When reading a circle from general form, completing the square wrongly — remember to subtract back the squared half-coefficients before reading r2.
  • Confusing the relations: ellipse uses c2 = a2 − b2, but hyperbola uses c2 = a2 + b2.
  • Forgetting that for an ellipse the larger denominator sits under the major-axis variable (it decides whether the major axis is along x or y).
  • Writing the wrong standard parabola: a y2 term ⇒ axis along x-axis; an x2 term ⇒ axis along y-axis. The sign of the coefficient fixes the direction it opens.
  • Using latus rectum = 4a for ellipses/hyperbolas — that is only for the parabola; for ellipse and hyperbola it is 2b2/a.
  • Taking the negative root for a, b, or r — lengths and semi-axes are always positive.

Practice MCQs & Assertion–Reason

1. The centre and radius of the circle x2 + y2 − 4x − 8y − 45 = 0 are:

(a) (2, 4), √65    (b) (−2, −4), √65    (c) (2, 4), 65    (d) (4, 8), √45

2. The focus of the parabola y2 = 12x is:

(a) (0, 3)    (b) (3, 0)    (c) (12, 0)    (d) (−3, 0)

3. The length of the latus rectum of the parabola x2 = −16y is:

(a) 4    (b) 8    (c) 16    (d) 32

4. For the ellipse x2/25 + y2/9 = 1, the eccentricity is:

(a) 3/5    (b) 4/5    (c) 5/4    (d) √5/3

5. The length of the latus rectum of the ellipse x2/16 + y2/9 = 1 is:

(a) 9/4    (b) 9/2    (c) 3/2    (d) 7/4

6. For an ellipse, the relation between a, b and c is:

(a) c2 = a2 + b2    (b) c2 = a2 − b2    (c) b2 = a2 + c2    (d) a2 = b2 − c2

7. The eccentricity of any hyperbola is:

(a) exactly 1    (b) less than 1    (c) greater than 1    (d) equal to 0

8. The vertices of the hyperbola x2/16 − y2/9 = 1 are:

(a) (±3, 0)    (b) (±4, 0)    (c) (±5, 0)    (d) (0, ±4)

9. The point (−2.5, 3.5) with respect to the circle x2 + y2 = 25 lies:

(a) inside    (b) outside    (c) on the circle    (d) at the centre

10. A conic is obtained as a section of a double-napped cone by a plane. When the plane is perpendicular to the axis (β = 90°), the section is a:

(a) parabola    (b) ellipse    (c) circle    (d) hyperbola

Answer key: 1-(a), 2-(b), 3-(c), 4-(b), 5-(b), 6-(b), 7-(c), 8-(b), 9-(a), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The eccentricity of an ellipse is always less than 1.

Reason: For an ellipse c < a, and the eccentricity is e = c/a.

A-R 2. Assertion: The latus rectum of the parabola y2 = 8x has length 8.

Reason: For y2 = 4ax the length of the latus rectum is 4a.

A-R 3. Assertion: For the hyperbola x2/16 − y2/9 = 1, the foci are (±5, 0).

Reason: In a hyperbola c2 = a2 − b2.

A-R 4. Assertion: The equation x2 + y2 + 4x − 6y − 3 = 0 represents a circle with centre (−2, 3).

Reason: For x2 + y2 + 2gx + 2fy + c = 0 the centre is (−g, −f).

A-R 5. Assertion: The eccentricity of a hyperbola can equal 1.

Reason: For a hyperbola c > a, so e = c/a > 1.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(D).

Quick Revision Summary

  • Circle: (x − h)2 + (y − k)2 = r2; from general form, centre (−g, −f), radius √(g2 + f2 − c).
  • Parabola y2 = 4ax: focus (a, 0), directrix x = −a, latus rectum 4a; the y2/x2 term and sign fix the axis and direction.
  • Ellipse x2/a2 + y2/b2 = 1 (a > b): c2 = a2 − b2, foci (±c, 0), e = c/a < 1, latus rectum 2b2/a.
  • Hyperbola x2/a2 − y2/b2 = 1: c2 = a2 + b2, foci (±c, 0), e = c/a > 1, latus rectum 2b2/a.
  • The major (ellipse) / transverse (hyperbola) axis is decided by which variable carries the larger/positive term.
  • Real-life models: parabolic reflectors and bridge cables use the parabola; semi-elliptical arches and orbit-type paths use the ellipse.

How to score full marks in this chapter

Always first identify the conic and write its standard equation, then read off a, b (and c) carefully. For ellipses and hyperbolas, decide the axis from the denominators before computing c, and keep eccentricity as e = c/a. Convert any equation like 16x2 + y2 = 16 to the standard /a2, /b2 form before extracting values. State the formula you use (latus rectum = 2b2/a, etc.) and present coordinates in pairs (±c, 0) so each result earns its mark.

Frequently Asked Questions

What is Class 11 Maths Chapter 10 Conic Sections about?

Chapter 10 studies the conic sections — circle, parabola, ellipse and hyperbola — obtained by cutting a double-napped cone with a plane. You learn their definitions, standard equations, and how to find centre, radius, focus, directrix, vertices, foci, eccentricity and latus rectum, plus real-life applications.

How many exercises are there in Class 11 Maths Chapter 10?

There are four numbered exercises — Exercise 10.1 (circle), 10.2 (parabola), 10.3 (ellipse) and 10.4 (hyperbola) — followed by a Miscellaneous Exercise. Every question of all five is solved step by step on this page.

What is the difference between the eccentricity of an ellipse and a hyperbola?

Both use e = c/a. For an ellipse c2 = a2 − b2 with c < a, so 0 < e < 1. For a hyperbola c2 = a2 + b2 with c > a, so e > 1.

Are these Class 11 Maths Chapter 10 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with answers verified against the book’s answer key.

← Chapter 9: Straight Lines Class 11 Maths Chapter 11: Three Dimensional Geometry →
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