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This page provides complete, step-by-step NCERT solutions for every exercise in Chapter 10 — Conic Sections. Each solution identifies the conic, states the relevant formula, substitutes values, simplifies, and gives the final answer. Try each problem before looking at the solution.
Key Concepts (Quick Recap)
- Circle: (x − h)² + (y − k)² = r²
- Parabola (standard): y² = 4ax → Focus (a, 0), Directrix x = −a, Latus rectum = 4a
- Parabola (vertical): x² = 4ay → Focus (0, a), Directrix y = −a, Latus rectum = 4a
- Ellipse: x²/a² + y²/b² = 1, where a > b, c² = a² − b², e = c/a
- Hyperbola: x²/a² − y²/b² = 1, c² = a² + b², e = c/a
- Latus rectum of ellipse: 2b²/a
- Latus rectum of hyperbola: 2b²/a
- Eccentricity: Circle e = 0, Parabola e = 1, Ellipse 0 < e < 1, Hyperbola e > 1
Exercise 10.1
Q1. Find the equation of the circle with centre (0, 2) and radius 2.
Using (x − h)² + (y − k)² = r²:
(x − 0)² + (y − 2)² = 4
x² + (y − 2)² = 4 (or x² + y² − 4y = 0)
Q2. Find the equation of the circle with centre (−2, 3) and radius 4.
(x + 2)² + (y − 3)² = 16
x² + 4x + 4 + y² − 6y + 9 = 16
x² + y² + 4x − 6y − 3 = 0
Q3. Find the centre and radius of the circle x² + y² + 6x − 4y − 12 = 0.
x² + 6x + y² − 4y = 12
(x + 3)² − 9 + (y − 2)² − 4 = 12
(x + 3)² + (y − 2)² = 25
Centre = (−3, 2), Radius = 5
Q4. Find the centre and radius of x² + y² − 4x − 8y − 5 = 0.
x² − 4x + y² − 8y = 5
(x − 2)² − 4 + (y − 4)² − 16 = 5
(x − 2)² + (y − 4)² = 25
Centre = (2, 4), Radius = 5
Q5. Find the equation of the circle with centre (1, 1) passing through (2, 3).
r² = (2 − 1)² + (3 − 1)² = 1 + 4 = 5
(x − 1)² + (y − 1)² = 5
x² + y² − 2x − 2y − 3 = 0
Q6. Find the equation of the circle passing through (0, 0), (2, 0) and (0, 3).
General equation: x² + y² + 2gx + 2fy + c = 0
Through (0, 0): c = 0.
Through (2, 0): 4 + 4g = 0 → g = −1.
Through (0, 3): 9 + 6f = 0 → f = −3/2.
Equation: x² + y² − 2x − 3y = 0
Centre = (1, 3/2), Radius = √(1 + 9/4) = √(13/4) = √13/2
Exercise 10.2
Q1. Find the coordinates of the focus, axis, equation of the directrix and latus rectum of the parabola y² = 12x.
Comparing with y² = 4ax: 4a = 12, so a = 3.
- Focus: (3, 0)
- Axis: x-axis (y = 0)
- Directrix: x = −3
- Latus rectum: 4a = 12
Q2. Find the focus, directrix and latus rectum of x² = 16y.
Comparing with x² = 4ay: 4a = 16, so a = 4.
- Focus: (0, 4)
- Axis: y-axis
- Directrix: y = −4
- Latus rectum: 4a = 16
Q3. Find the equation of the parabola with focus (6, 0) and directrix x = −6.
Focus at (6, 0), directrix x = −6. Vertex is the midpoint: (0, 0).
4a = 2 × 6 = 12, so a = 6.
Equation: y² = 24x
Q4. Find the equation of the parabola with focus (0, −3) and directrix y = 3.
Focus at (0, −3), directrix y = 3. Vertex at (0, 0).
4a = 2 × 3 = 6, so a = 3. The parabola opens downward.
Equation: x² = −12y
Q5. Find the focus and latus rectum of y² = −8x.
4a = 8, a = 2. Parabola opens left.
Focus: (−2, 0), Latus rectum = 8
Q6. Find the equation of the parabola with vertex at (0, 0) and focus at (5, 0).
Focus at (5, 0), vertex at (0, 0). Opens right, a = 5.
Equation: y² = 20x
Q7. Find the equation of the parabola with vertex at (0, 0) and focus at (0, −4).
Opens downward, a = 4.
Equation: x² = −16y
Q8. Find the focus and latus rectum of x² = −20y.
4a = 20, a = 5. Opens downward.
Focus: (0, −5), Latus rectum = 20
Exercise 10.3
Q1. Find the coordinates of the foci, vertices, and the length of the major axis of the ellipse x²/25 + y²/9 = 1.
a² = 25, b² = 9, so a = 5, b = 3.
c² = a² − b² = 25 − 9 = 16, c = 4.
- Foci: (±4, 0)
- Vertices: (±5, 0)
- Major axis length: 2a = 10
- Minor axis length: 2b = 6
- Eccentricity: e = 4/5
Q2. Find the foci, vertices and eccentricity of x²/16 + y²/9 = 1.
a = 4, b = 3, c = √(16 − 9) = √7.
- Foci: (±√7, 0)
- Vertices: (±4, 0)
- Eccentricity: e = √7/4
Q3. Find the equation of the ellipse with foci (±3, 0) and vertices (±5, 0).
a = 5, c = 3, b² = a² − c² = 25 − 9 = 16, b = 4.
Equation: x²/25 + y²/16 = 1
Q4. Find the equation of the ellipse with foci (0, ±4) and vertices (0, ±6).
a = 6, c = 4, b² = 36 − 16 = 20.
Equation: x²/20 + y²/36 = 1
Q5. Find the foci, vertices and eccentricity of x²/9 + y²/36 = 1.
Here a² = 36 (major axis along y), b² = 9. a = 6, b = 3.
c² = 36 − 9 = 27, c = 3√3.
- Foci: (0, ±3√3)
- Vertices: (0, ±6)
- Eccentricity: e = 3√3/6 = √3/2
Q6. Find the equation of the ellipse with centre at origin, major axis along x-axis, a = 5 and passing through (3, 12/5).
Equation: x²/25 + y²/b² = 1.
Passes through (3, 12/5): 9/25 + 144/(25b²) = 1
144/(25b²) = 1 − 9/25 = 16/25
b² = 144/16 = 9, b = 3.
Equation: x²/25 + y²/9 = 1
Q7. Find the equation of the ellipse with eccentricity 1/2 and foci (±2, 0).
c = 2, e = 1/2 = c/a → a = 4.
b² = a² − c² = 16 − 4 = 12.
Equation: x²/16 + y²/12 = 1
Q8. Find the foci and latus rectum of x²/100 + y²/36 = 1.
a = 10, b = 6, c = √(100 − 36) = √64 = 8.
- Foci: (±8, 0)
- Latus rectum = 2b²/a = 2(36)/10 = 36/5
Q9. Find the equation of the ellipse with vertices (±6, 0) and foci (±4, 0).
a = 6, c = 4, b² = 36 − 16 = 20.
Equation: x²/36 + y²/20 = 1
Q10. Find the latus rectum of the ellipse x²/36 + y²/16 = 1.
a = 6, b = 4.
Latus rectum = 2b²/a = 2(16)/6 = 16/3
Exercise 10.4
Q1. Find the coordinates of the foci, vertices and eccentricity of the hyperbola x²/16 − y²/9 = 1.
a² = 16, b² = 9, a = 4, b = 3.
c² = a² + b² = 16 + 9 = 25, c = 5.
- Foci: (±5, 0)
- Vertices: (±4, 0)
- Eccentricity: e = 5/4
Q2. Find the foci and eccentricity of x²/9 − y²/4 = 1.
a = 3, b = 2, c = √(9 + 4) = √13.
- Foci: (±√13, 0)
- Eccentricity: e = √13/3
Q3. Find the equation of the hyperbola with foci (±5, 0) and vertices (±3, 0).
a = 3, c = 5, b² = c² − a² = 25 − 9 = 16.
Equation: x²/9 − y²/16 = 1
Q4. Find the equation of the hyperbola with foci (0, ±13) and vertices (0, ±5).
a = 5, c = 13, b² = 169 − 25 = 144.
Equation: y²/25 − x²/144 = 1
Q5. Find the foci, vertices and latus rectum of y²/16 − x²/9 = 1.
a² = 16 (along y), b² = 9. a = 4, b = 3.
c² = 16 + 9 = 25, c = 5.
- Foci: (0, ±5)
- Vertices: (0, ±4)
- Latus rectum = 2b²/a = 18/4 = 9/2
Q6. Find the equation of the hyperbola with vertices (±7, 0) and e = 4/3.
a = 7, e = 4/3. c = ae = 28/3.
b² = c² − a² = 784/9 − 49 = 784/9 − 441/9 = 343/9.
Equation: x²/49 − 9y²/343 = 1 (or x²/49 − y²/(343/9) = 1)
Q7. Find the equation of the hyperbola with foci (0, ±√10) and passing through (2, 3).
c² = 10, equation: y²/a² − x²/b² = 1 where a² + b² = 10.
Passes through (2, 3): 9/a² − 4/b² = 1.
Let a² = t. Then b² = 10 − t.
9/t − 4/(10 − t) = 1
9(10 − t) − 4t = t(10 − t)
90 − 9t − 4t = 10t − t²
90 − 13t = 10t − t²
t² − 23t + 90 = 0
(t − 5)(t − 18) = 0 → t = 5 or t = 18.
Since a² < c² = 10, a² = 5, b² = 5.
Equation: y²/5 − x²/5 = 1 (or y² − x² = 5)
Q8. Find the eccentricity of the hyperbola with latus rectum = 8 and conjugate axis = 2√2.
Conjugate axis = 2b = 2√2, so b = √2, b² = 2.
Latus rectum = 2b²/a = 8 → 4/a = 8 → a = 1/2, a² = 1/4.
c² = a² + b² = 1/4 + 2 = 9/4, c = 3/2.
e = c/a = (3/2)/(1/2) = 3
Miscellaneous Exercise
Q1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Let the parabola be y² = 4ax. The rim point is (5, 10) (half the diameter = 10).
100 = 4a(5) → a = 5.
Focus is at (5, 0), which is 5 cm from the vertex.
Q2. 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.
Equation: x²/16 + y²/4 = 1 (a = 4, b = 2).
Point 1.5 m from one end means x = 4 − 1.5 = 2.5 m from centre (or x = −2.5 + 8 = measured from the other end).
Actually, 1.5 m from one end (at x = −4) means x = −4 + 1.5 = −2.5.
y² = 4(1 − x²/16) = 4(1 − 6.25/16) = 4(9.75/16) = 39/16
y = √(39)/4 ≈ 1.56 m
Q3. A rod of length 12 cm moves with its ends always touching the coordinate axes. Find the equation of the locus of a point P on the rod at a distance of 3 cm from the end in contact with the x-axis.
Let the end on x-axis be (a, 0) and on y-axis be (0, b). Then a² + b² = 144.
Point P divides the rod in ratio 3 : 9 = 1 : 3 from the x-axis end.
P = ((3/12) × 0 + (9/12) × a, (3/12) × b + (9/12) × 0) = (3a/4, b/4).
Wait — P is 3 cm from the x-axis end, so it divides in ratio 3 : 9.
h = (9 × 0 + 3 × a)/12 = a/4, no — let me redo.
P is at distance 3 from end on x-axis. Coordinates of P:
x = (9/12)(a) + (3/12)(0) = 3a/4... Actually:
End A is on x-axis: (a, 0). End B is on y-axis: (0, b). P is 3 cm from A.
P = A + (3/12)(B − A) = (a, 0) + (1/4)(−a, b) = (a − a/4, b/4) = (3a/4, b/4).
Let P = (h, k). Then h = 3a/4 → a = 4h/3, k = b/4 → b = 4k.
Substituting in a² + b² = 144: (4h/3)² + (4k)² = 144
16h²/9 + 16k² = 144 → h²/9 + k² = 9 → h²/81 + k²/9 = 1
Equation of locus: x²/81 + y²/9 = 1 (an ellipse)
Q4. Find the area enclosed by the ellipse x²/a² + y²/b² = 1.
Area = πab. (Standard result — the area of an ellipse with semi-axes a and b.)
Area = πab
Q5. Find the equation of the circle passing through the foci of the ellipse x²/25 + y²/9 = 1 and with centre at (0, 0).
Foci: c = √(25 − 9) = 4, so foci at (±4, 0).
Circle passes through (4, 0) and (−4, 0) with centre (0, 0).
r = 4, equation: x² + y² = 16
Q6. Find the equation of the hyperbola with foci (±6, 0) and e = 3/2.
c = 6, e = 3/2. a = c/e = 6/(3/2) = 4.
b² = c² − a² = 36 − 16 = 20.
Equation: x²/16 − y²/20 = 1
Q7. 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 path traced by the man.
Sum of distances = 10 = 2a, so a = 5. Distance between foci = 8 = 2c, so c = 4.
b² = a² − c² = 25 − 16 = 9.
Path is an ellipse: x²/25 + y²/9 = 1
Q8. Find the equation of the hyperbola where foci are (0, ±12) and latus rectum is 36.
c = 12, latus rectum = 2b²/a = 36, so b² = 18a.
Also c² = a² + b² → 144 = a² + 18a
a² + 18a − 144 = 0
a = [−18 ± √(324 + 576)]/2 = [−18 ± √900]/2 = [−18 ± 30]/2
a = 6 (taking positive value).
b² = 18(6) = 108.
Equation: y²/36 − x²/108 = 1
Q9. Find the equation of the ellipse with centre at (0, 0), major axis on the y-axis and passing through (1, 3) and (2, 2).
Equation: x²/b² + y²/a² = 1.
Through (1, 3): 1/b² + 9/a² = 1 …(1)
Through (2, 2): 4/b² + 4/a² = 1 …(2)
From (2): 4/b² = 1 − 4/a² → 1/b² = (1 − 4/a²)/4.
Substituting in (1): (1 − 4/a²)/4 + 9/a² = 1
1/4 − 1/a² + 9/a² = 1
8/a² = 3/4
a² = 32/3
1/b² = (1 − 4/(32/3))/4 = (1 − 3/8)/4 = (5/8)/4 = 5/32
b² = 32/5
Equation: 5x²/32 + 3y²/32 = 1 (or 5x² + 3y² = 32)
Q10. Find the eccentricity of the ellipse x²/9 + y²/4 = 1.
a = 3, b = 2. c = √(9 − 4) = √5.
e = √5/3 = √5/3 ≈ 0.745
Internal Links
- [Chapter 10 Guide — Conic Sections](/school/cbse/class/11/maths/cbse-conic-sections)
- [All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
- [Class 11 Maths Subject Guide](/school/cbse/class/11/maths)