CBSE Important Questions Class 11 Maths Chapter 10 – Conic Sections

Class 11 Maths Chapter 10 Conic Sections is part of the CBSE Class 11 Maths syllabus. A conic section is the curve we get when a plane cuts a double cone. Students learn the equations and properties of the circle, the parabola, the ellipse and the hyperbola, along with terms such as focus, directrix, vertex, axis, latus rectum and eccentricity.

These CBSE important questions for Class 11 Maths Chapter 10 cover finding the centre and radius of a circle, the equation of a circle from given conditions, the focus and latus rectum of a parabola, the foci and eccentricity of an ellipse and a hyperbola, and converting a general second-degree equation into standard form. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Find the centre and radius of the circle x² + y² − 4x − 8y − 45 = 0.

Answer: Complete the square for the x terms and the y terms:
(x² − 4x) + (y² − 8y) = 45
(x² − 4x + 4) + (y² − 8y + 16) = 45 + 4 + 16
(x − 2)² + (y − 4)² = 65
Comparing with (x − h)² + (y − k)² = r²:
Centre = (2, 4) and radius = √65 units

Q2. Find the equation of the circle with centre (−3, 2) and radius 4.

Answer: The standard equation of a circle is
(x − h)² + (y − k)² = r²
Here h = −3, k = 2 and r = 4.
(x + 3)² + (y − 2)² = 16
Expanding:
x² + 6x + 9 + y² − 4y + 4 = 16
x² + y² + 6x − 4y − 3 = 0

Q3. Find the coordinates of the focus, the axis, the equation of the directrix and the length of the latus rectum of the parabola y² = 12x.

Answer: Compare with the standard form y² = 4ax:
4a = 12, so a = 3
• Focus: (a, 0) = (3, 0)
• Axis: the x-axis (that is y = 0)
• Directrix: x = −a, that is x = −3
• Length of latus rectum: 4a = 12 units
Since the equation is y² = 4ax with a positive, the parabola opens to the right.

Q4. Find the equation of the parabola with vertex at the origin, axis along the y-axis and passing through the point (5, 2).

Answer: With the vertex at the origin and the axis along the y-axis, the equation is of the form
x² = 4ay
The point (5, 2) lies on it, so:
5² = 4a(2)
25 = 8a
a = 25/8
So the equation is x² = 4(25/8)y
x² = (25/2)y, or 2x² = 25y

Q5. Find the foci, the vertices, the eccentricity and the length of the latus rectum of the ellipse x²/25 + y²/9 = 1.

Answer: Here a² = 25 and b² = 9, so a = 5 and b = 3. Since a > b, the major axis is along the x-axis.
c² = a² − b² = 25 − 9 = 16, so c = 4
• Foci: (±c, 0) = (±4, 0)
• Vertices: (±a, 0) = (±5, 0)
• Eccentricity: e = c/a = 4/5
• Latus rectum: 2b²/a = 2(9)/5 = 18/5 units

Q6. Find the foci, the vertices, the eccentricity and the length of the latus rectum of the hyperbola x²/9 − y²/16 = 1.

Answer: Here a² = 9 and b² = 16, so a = 3 and b = 4.
For a hyperbola, c² = a² + b² = 9 + 16 = 25, so c = 5
• Foci: (±5, 0)
• Vertices: (±3, 0)
• Eccentricity: e = c/a = 5/3
• Latus rectum: 2b²/a = 2(16)/3 = 32/3 units
Note that for a hyperbola e is always greater than 1, while for an ellipse it is less than 1.

Q7. Find the equation of the ellipse whose vertices are (±13, 0) and foci are (±5, 0).

Answer: Since the vertices and foci lie on the x-axis, the major axis is along the x-axis and the form is
x²/a² + y²/b² = 1
From the vertices, a = 13, so a² = 169.
From the foci, c = 5, so c² = 25.
Using b² = a² − c²:
b² = 169 − 25 = 144
So the equation is
x²/169 + y²/144 = 1

Q8. Find the equation of the circle passing through the points (2, 3) and (−1, 1) whose centre lies on the line x − 3y − 11 = 0.

Answer: Let the centre be (h, k). Since it lies on the line:
h − 3k − 11 = 0 ...(i)
The distances from the centre to both points are equal (both are radii):
(h − 2)² + (k − 3)² = (h + 1)² + (k − 1)²
h² − 4h + 4 + k² − 6k + 9 = h² + 2h + 1 + k² − 2k + 1
−6h − 4k + 11 = 0 ...(ii)
Solving (i) and (ii) gives h = 7/2 and k = −5/2.
Radius² = (7/2 − 2)² + (−5/2 − 3)² = 9/4 + 121/4 = 130/4
So the equation is
(x − 7/2)² + (y + 5/2)² = 65/2

Q9. Define a conic section. Name the four types of conic sections and state the eccentricity of each.

Answer: A conic section is the curve formed when a plane cuts a double right circular cone. The shape depends on the angle at which the plane cuts the cone.

Conic Eccentricity (e)
Circle e = 0
Ellipse 0 < e < 1
Parabola e = 1
Hyperbola e > 1

Q10. Find the equation of the parabola with focus (6, 0) and directrix x = −6.

Answer: The focus lies on the positive x-axis and the directrix is a vertical line, so the parabola opens to the right and its vertex is at the origin.
The standard form is y² = 4ax, where the focus is (a, 0).
Here a = 6.
So the equation is
y² = 4(6)x
y² = 24x

Q11. Find the equation of the hyperbola with foci (0, ±13) and the length of the conjugate axis 24.

Answer: Since the foci lie on the y-axis, the form is
y²/a² − x²/b² = 1
From the foci, c = 13.
The conjugate axis has length 2b = 24, so b = 12 and b² = 144.
Using c² = a² + b²:
169 = a² + 144
a² = 25
So the equation is
y²/25 − x²/144 = 1

Q12. 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.

Answer: Take the centre of the base as the origin. The semi-major axis a = 8/2 = 4 and the semi-minor axis b = 2.
The equation is x²/16 + y²/4 = 1.
A point 1.5 m from one end is at a distance 4 − 1.5 = 2.5 m from the centre, so x = 2.5.
Substituting:
(2.5)²/16 + y²/4 = 1
6.25/16 + y²/4 = 1
y²/4 = 1 − 0.390625 = 0.609375
y² = 2.4375
y = 1.56 m (approximately)

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