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NCERT Solutions · Class 11 Maths · Chapter 10

Chapter 10: Conic Sections (Coordinate Geometry)

Step-by-step solutions to all exercises of Chapter 10, Conic Sections (NCERT Class 11 Maths, 2026-27): circles, parabolas, ellipses and hyperbolas: centre and radius, focus, directrix, latus rectum, eccentricity and equations from given data. All 70 questions are answered, with the key answer highlighted.

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Circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Parabola $y^{2} = 4 a x$ : focus (a, 0), directrix x = −a, latus rectum 4a. Ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ : $c^{2} = a^{2} - b^{2}$ , $e = \frac{c}{a}$ , latus rectum $\frac{2 b ^{2}}{a}$ . Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ : $c^{2} = a^{2} + b^{2}$ .

Exercise 10.1

Find the equation of the circle with: centre (0, 2), radius 2

Solution

x² + (y − 2)² = 4 ⇒ x² + y² − 4y = 0.

x² + y² − 4y = 0

Find the equation of the circle with: centre (−2, 3), radius 4

Solution

(x + 2)² + (y − 3)² = 16 ⇒ x² + y² + 4x − 6y − 3 = 0.

x² + y² + 4x − 6y − 3 = 0

Find the equation of the circle with: centre (½, ¼), radius 1/12

Solution

(x − ½)² + (y − ¼)² = 1/144 ⇒ x² + y² − x − y/2 + 1/4 + 1/16 − 1/144 = 0; × 144 and ÷ 4: 36x² + 36y² − 36x − 18y + 11 = 0.

36x² + 36y² − 36x − 18y + 11 = 0

Find the equation of the circle with: centre (1, 1), radius √2

Solution

(x − 1)² + (y − 1)² = 2 ⇒ x² + y² − 2x − 2y = 0.

x² + y² − 2x − 2y = 0

Find the equation of the circle with: centre (−a, −b), radius √(a² − b²).

Solution

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

x² + y² + 2ax + 2by + 2b² = 0

Find the centre and radius: (x + 5)² + (y − 3)² = 36

Solution

centre (−5, 3), r = 6.

centre (−5, 3), r = 6

Find the centre and radius: x² + y² − 4x − 8y − 45 = 0

Solution

(x − 2)² + (y − 4)² = 45 + 4 + 16 = 65: centre (2, 4), r = √65.

centre (2, 4), r = √65

Find the centre and radius: x² + y² − 8x + 10y − 12 = 0

Solution

(x − 4)² + (y + 5)² = 12 + 16 + 25 = 53: centre (4, −5), r = √53.

centre (4, −5), r = √53

Find the centre and radius: 2x² + 2y² − x = 0.

Solution

x² − x/2 + y² = 0 ⇒ (x − ¼)² + y² = 1/16: centre (¼, 0), r = ¼.

centre (¼, 0), r = ¼

Circle through (4, 1) and (6, 5) with centre on 4x + y = 16.

Solution

Equal distances from (h, k): (4 − h)² + (1 − k)² = (6 − h)² + (5 − k)² ⇒ 4h + 8k = 44 ⇒ h + 2k = 11. With 4h + k = 16: h = 3, k = 4; r² = 1 + 9 = 10. (x − 3)² + (y − 4)² = 10 ⇒ x² + y² − 6x − 8y + 15 = 0.

x² + y² − 6x − 8y + 15 = 0

Circle through (2, 3) and (−1, 1) with centre on x − 3y − 11 = 0.

Solution

(2 − h)² + (3 − k)² = (1 + h)² + (1 − k)² ⇒ 6h + 4k = 11. With h − 3k = 11: h = 7/2, k = −5/2. r² = (3/2)² + (11/2)² = 130/4. Expanding: x² + y² − 7x + 5y − 14 = 0.

x² + y² − 7x + 5y − 14 = 0

Circle of radius 5 with centre on the x-axis through (2, 3).

Solution

(2 − h)² + 9 = 25 ⇒ h = −2 or 6: x² + y² + 4x − 21 = 0 or x² + y² − 12x + 11 = 0.

x² + y² + 4x − 21 = 0; x² + y² − 12x + 11 = 0

Circle through (0, 0) making intercepts a and b on the axes.

Solution

It passes through (0, 0), (a, 0), (0, b); the right angle at O means (a, 0)–(0, b) is a diameter: centre (a/2, b/2): x² + y² − ax − by = 0.

x² + y² − ax − by = 0

Circle with centre (2, 2) through (4, 5).

Solution

r² = 4 + 9 = 13: (x − 2)² + (y − 2)² = 13 ⇒ x² + y² − 4x − 4y = 5.

x² + y² − 4x − 4y = 5

Is (−2.5, 3.5) inside, outside or on x² + y² = 25?

Solution

Distance² from O = 6.25 + 12.25 = 18.5 < 25: inside.

inside

Exercise 10.2

Find the focus, axis, directrix and latus rectum: y² = 12x

Solution

Compare with 4a. a = 3: F(3, 0), x-axis, x = −3, LR 12.

F(3, 0), x-axis, x = −3, LR 12

Find the focus, axis, directrix and latus rectum: x² = 6y

Solution

Compare with 4a. a = 3/2: F(0, 3/2), y-axis, y = −3/2, LR 6.

F(0, 3/2), y-axis, y = −3/2, LR 6

Find the focus, axis, directrix and latus rectum: y² = −8x

Solution

Compare with 4a. a = 2: F(−2, 0), x-axis, x = 2, LR 8.

F(−2, 0), x-axis, x = 2, LR 8

Find the focus, axis, directrix and latus rectum: x² = −16y

Solution

Compare with 4a. a = 4: F(0, −4), y-axis, y = 4, LR 16.

F(0, −4), y-axis, y = 4, LR 16

Find the focus, axis, directrix and latus rectum: y² = 10x

Solution

Compare with 4a. a = 5/2: F(5/2, 0), x-axis, x = −5/2, LR 10.

F(5/2, 0), x-axis, x = −5/2, LR 10

Find the focus, axis, directrix and latus rectum: x² = −9y.

Solution

Compare with 4a. a = 9/4: F(0, −9/4), y-axis, y = 9/4, LR 9.

F(0, −9/4), y-axis, y = 9/4, LR 9

Find the parabola: focus (6, 0), directrix x = −6

Solution

a = 6: y² = 24x.

y² = 24x

Find the parabola: focus (0, −3), directrix y = 3

Solution

opens down, a = 3: x² = −12y.

x² = −12y

Find the parabola: vertex O, focus (3, 0)

Solution

y² = 12x.

y² = 12x

Find the parabola: vertex O, focus (−2, 0)

Solution

y² = −8x.

y² = −8x

Find the parabola: vertex O, through (2, 3), axis the x-axis

Solution

y² = 4ax: 9 = 8a ⇒ a = 9/8: y² = 9x/2 ⇒ 2y² = 9x.

2y² = 9x

Find the parabola: vertex O, through (5, 2), symmetric about the y-axis.

Solution

x² = 4ay: 25 = 8a: x² = 25y/2 ⇒ 2x² = 25y.

2x² = 25y

Exercise 10.3

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: x²/36 + y²/16 = 1

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. a = 6, b = 4, c = 2√5: F(±2√5, 0), V(±6, 0), 12, 8, e = √5/3, LR 16/3.

F(±2√5, 0), V(±6, 0), 12, 8, e = √5/3, LR 16/3

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: x²/4 + y²/25 = 1

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. a = 5 (y), b = 2, c = √21: F(0, ±√21), V(0, ±5), 10, 4, e = √21/5, LR 8/5.

F(0, ±√21), V(0, ±5), 10, 4, e = √21/5, LR 8/5

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: x²/16 + y²/9 = 1

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. a = 4, b = 3, c = √7: F(±√7, 0), V(±4, 0), 8, 6, e = √7/4, LR 9/2.

F(±√7, 0), V(±4, 0), 8, 6, e = √7/4, LR 9/2

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: x²/25 + y²/100 = 1

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. a = 10 (y), b = 5, c = 5√3: F(0, ±5√3), V(0, ±10), 20, 10, e = √3/2, LR 5.

F(0, ±5√3), V(0, ±10), 20, 10, e = √3/2, LR 5

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: x²/49 + y²/36 = 1

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. a = 7, b = 6, c = √13: F(±√13, 0), V(±7, 0), 14, 12, e = √13/7, LR 72/7.

F(±√13, 0), V(±7, 0), 14, 12, e = √13/7, LR 72/7

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: x²/100 + y²/400 = 1

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. a = 20 (y), b = 10, c = 10√3: F(0, ±10√3), V(0, ±20), 40, 20, e = √3/2, LR 10.

F(0, ±10√3), V(0, ±20), 40, 20, e = √3/2, LR 10

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: 36x² + 4y² = 144

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. ÷144: x²/4 + y²/36 = 1: a = 6 (y), b = 2, c = 4√2: F(0, ±4√2), V(0, ±6), 12, 4, e = 2√2/3, LR 4/3.

F(0, ±4√2), V(0, ±6), 12, 4, e = 2√2/3, LR 4/3

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: 16x² + y² = 16

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. x²/1 + y²/16 = 1: a = 4 (y), b = 1, c = √15: F(0, ±√15), V(0, ±4), 8, 2, e = √15/4, LR 1/2.

F(0, ±√15), V(0, ±4), 8, 2, e = √15/4, LR 1/2

Find the foci, vertices, major and minor axes, eccentricity and latus rectum: 4x² + 9y² = 36.

Solution

Use c = √(a² − b²), e = c/a, LR = 2b²/a. x²/9 + y²/4 = 1: a = 3, b = 2, c = √5: F(±√5, 0), V(±3, 0), 6, 4, e = √5/3, LR 8/3.

F(±√5, 0), V(±3, 0), 6, 4, e = √5/3, LR 8/3

Find the ellipse: vertices (±5, 0), foci (±4, 0)

Solution

b² = a² − c² (or a² = b² + c²). b² = 9: x²/25 + y²/9 = 1.

x²/25 + y²/9 = 1

Find the ellipse: vertices (0, ±13), foci (0, ±5)

Solution

b² = a² − c² (or a² = b² + c²). b² = 144: x²/144 + y²/169 = 1.

x²/144 + y²/169 = 1

Find the ellipse: vertices (±6, 0), foci (±4, 0)

Solution

b² = a² − c² (or a² = b² + c²). b² = 20: x²/36 + y²/20 = 1.

x²/36 + y²/20 = 1

Find the ellipse: ends of major axis (±3, 0), minor (0, ±2)

Solution

b² = a² − c² (or a² = b² + c²). x²/9 + y²/4 = 1.

x²/9 + y²/4 = 1

Find the ellipse: ends of major axis (0, ±√5), minor (±1, 0)

Solution

b² = a² − c² (or a² = b² + c²). x²/1 + y²/5 = 1.

x²/1 + y²/5 = 1

Find the ellipse: major axis 26, foci (±5, 0)

Solution

b² = a² − c² (or a² = b² + c²). a = 13, b² = 144: x²/169 + y²/144 = 1.

x²/169 + y²/144 = 1

Find the ellipse: minor axis 16, foci (0, ±6)

Solution

b² = a² − c² (or a² = b² + c²). b = 8, a² = 64 + 36 = 100: x²/64 + y²/100 = 1.

x²/64 + y²/100 = 1

Find the ellipse: foci (±3, 0), a = 4

Solution

b² = a² − c² (or a² = b² + c²). b² = 16 − 9 = 7: x²/16 + y²/7 = 1.

x²/16 + y²/7 = 1

Find the ellipse: b = 3, c = 4, centre O, foci on the x-axis.

Solution

b² = a² − c² (or a² = b² + c²). a² = 9 + 16 = 25: x²/25 + y²/9 = 1.

x²/25 + y²/9 = 1

Centre O, major axis on the y-axis, through (3, 2) and (1, 6).

Solution

x²/b² + y²/a² = 1; put u = 1/b², v = 1/a²: 9u + 4v = 1, u + 36v = 1 ⇒ u = 1 − 36v ⇒ 9 − 320v = 1 ⇒ v = 1/40, u = 1/10: x²/10 + y²/40 = 1.

x²/10 + y²/40 = 1

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

Solution

16u + 9v = 1, 36u + 4v = 1 (u = 1/a², v = 1/b²): ×4 and ×9 then subtract: 260u = 5 ⇒ u = 1/52, v = 1/13: x²/52 + y²/13 = 1 (x² + 4y² = 52).

x²/52 + y²/13 = 1

Exercise 10.4

Find the foci, vertices, eccentricity and latus rectum: x²/16 − y²/9 = 1

Solution

Use c = √(a² + b²), e = c/a, LR = 2b²/a. a = 4, b = 3, c = 5: F(±5, 0), V(±4, 0), e = 5/4, LR 9/2.

F(±5, 0), V(±4, 0), e = 5/4, LR 9/2

Find the foci, vertices, eccentricity and latus rectum: y²/9 − x²/27 = 1

Solution

Use c = √(a² + b²), e = c/a, LR = 2b²/a. a = 3 (y), b² = 27, c = 6: F(0, ±6), V(0, ±3), e = 2, LR 18.

F(0, ±6), V(0, ±3), e = 2, LR 18

Find the foci, vertices, eccentricity and latus rectum: 9y² − 4x² = 36

Solution

Use c = √(a² + b²), e = c/a, LR = 2b²/a. ÷36: y²/4 − x²/9 = 1: a = 2, b = 3, c = √13: F(0, ±√13), V(0, ±2), e = √13/2, LR 9.

F(0, ±√13), V(0, ±2), e = √13/2, LR 9

Find the foci, vertices, eccentricity and latus rectum: 16x² − 9y² = 576

Solution

Use c = √(a² + b²), e = c/a, LR = 2b²/a. ÷576: x²/36 − y²/64 = 1: a = 6, b = 8, c = 10: F(±10, 0), V(±6, 0), e = 5/3, LR 64/3.

F(±10, 0), V(±6, 0), e = 5/3, LR 64/3

Find the foci, vertices, eccentricity and latus rectum: 5y² − 9x² = 36

Solution

Use c = √(a² + b²), e = c/a, LR = 2b²/a. ÷36: y²/(36/5) − x²/4 = 1: a = 6/√5, b = 2, c² = 56/5: F(0, ±2√14/√5), V(0, ±6/√5), e = √14/3, LR 4√5/3.

F(0, ±2√14/√5), V(0, ±6/√5), e = √14/3, LR 4√5/3

Find the foci, vertices, eccentricity and latus rectum: 49y² − 16x² = 784.

Solution

Use c = √(a² + b²), e = c/a, LR = 2b²/a. ÷784: y²/16 − x²/49 = 1: a = 4, b = 7, c = √65: F(0, ±√65), V(0, ±4), e = √65/4, LR 49/2.

F(0, ±√65), V(0, ±4), e = √65/4, LR 49/2

Find the hyperbola: vertices (±2, 0), foci (±3, 0)

Solution

b² = c² − a². b² = 5: x²/4 − y²/5 = 1.

x²/4 − y²/5 = 1

Find the hyperbola: vertices (0, ±5), foci (0, ±8)

Solution

b² = c² − a². b² = 39: y²/25 − x²/39 = 1.

y²/25 − x²/39 = 1

Find the hyperbola: vertices (0, ±3), foci (0, ±5)

Solution

b² = c² − a². b² = 16: y²/9 − x²/16 = 1.

y²/9 − x²/16 = 1

Find the hyperbola: foci (±5, 0), transverse axis 8

Solution

b² = c² − a². a = 4, b² = 9: x²/16 − y²/9 = 1.

x²/16 − y²/9 = 1

Find the hyperbola: foci (0, ±13), conjugate axis 24.

Solution

b² = c² − a². b = 12, a² = 169 − 144 = 25: y²/25 − x²/144 = 1.

y²/25 − x²/144 = 1

Foci (±3√5, 0), latus rectum 8.

Solution

2b²/a = 8 ⇒ b² = 4a; a² + 4a = 45 ⇒ (a + 9)(a − 5) = 0 ⇒ a = 5, b² = 20: x²/25 − y²/20 = 1.

x²/25 − y²/20 = 1

Foci (±4, 0), latus rectum 12.

Solution

b² = 6a; a² + 6a = 16 ⇒ a = 2, b² = 12: x²/4 − y²/12 = 1.

x²/4 − y²/12 = 1

Vertices (±7, 0), e = 4/3.

Solution

c = ae = 28/3; b² = 784/9 − 49 = 343/9: x²/49 − 9y²/343 = 1.

x²/49 − 9y²/343 = 1

Foci (0, ±√10), through (2, 3).

Solution

y²/a² − x²/b² = 1 with a² + b² = 10: 9/a² − 4/(10 − a²) = 1 ⇒ a⁴ − 23a² + 90 = 0 ⇒ a² = 5 or 18. a² = 18 gives b² < 0, rejected. a² = b² = 5: y²/5 − x²/5 = 1.

y²/5 − x²/5 = 1

Miscellaneous Exercise on Chapter 10

A parabolic reflector is 20 cm in diameter and 5 cm deep. Find the focus.

Solution

Vertex at O, axis along x: y² = 4ax through (5, 10): 100 = 20a ⇒ a = 5. Focus (5, 0): at the midpoint of the given diameter (5 cm from the vertex).

at the midpoint of the given diameter

A parabolic arch with vertical axis is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex?

Solution

Vertex at the top: x² = −4ay through (2.5, −10): 6.25 = 40a ⇒ a = 5/32. At y = −2: x² = 4(5/32)(2) = 5/4 ⇒ x = √5/2. Width 2x = √5 ≈ 2.23 m.

√5 ≈ 2.23 m

A suspension-bridge cable is a parabola; the roadway is 100 m long, the longest wire 30 m, the shortest 6 m. Find the wire 18 m from the middle.

Solution

Lowest point of the cable (0, 6) above the road's middle: x² = 4a(y − 6) through (50, 30): 2500 = 96a ⇒ 4a = 2500/24. At x = 18: y − 6 = 324 × 24/2500 = 3.11 ⇒ y ≈ 9.11 m.

y ≈ 9.11 m

A semi-elliptical arch is 8 m wide and 2 m high at the centre. Find the height 1.5 m from one end.

Solution

x²/16 + y²/4 = 1; 1.5 m from an end is x = 2.5: y² = 4(1 − 6.25/16) = 2.4375 ⇒ y ≈ 1.56 m.

y ≈ 1.56 m

A rod of length 12 cm moves with its ends on the axes. Find the locus of P on the rod 3 cm from the end on the x-axis.

Solution

As in Example 19 with AP = 3, PB = 9: x = 9 cos θ, y = 3 sin θ ⇒ x²/81 + y²/9 = 1.

x²/81 + y²/9 = 1

Area of the triangle joining the vertex of x² = 12y to the ends of its latus rectum.

Solution

a = 3: latus rectum ends (±6, 3). Area = ½ × 12 × 3 = 18 square units.

18 square units

A runner's distances from two flag posts 8 m apart always add to 10 m. Find his path.

Solution

Ellipse with 2a = 10, 2c = 8: b² = 25 − 16 = 9: x²/25 + y²/9 = 1.

x²/25 + y²/9 = 1

An equilateral triangle is inscribed in y² = 4ax with one vertex at the vertex of the parabola. Find its side.

Solution

By symmetry the other vertices are (h, ±h tan 30°) = (h, ±h/√3). On the parabola: h²/3 = 4ah ⇒ h = 12a. Side = 2h/√3 = 24a/√3 = 8√3a.

8√3a

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