Class 11 Maths Chapter 10: Conic Sections MCQs with answers

Answer: (c) an ellipse

When the plane cuts one nappe with α < β < 90°, the section is an ellipse. β = 90° gives a circle, β = α a parabola, and 0 ≤ β < α a hyperbola.

Answer: (d) a straight line

At the vertex with β = α the plane contains a generator and the section is a straight line, the degenerated case of a parabola. α < β ≤ 90° gives a point; 0 ≤ β < α a pair of intersecting lines.

Answer: (a) 16

g = −3, f = 4, so the radius is √(g² + f² − k) = √(9 + 16 − k). Then 25 − k = 9 and k = 16.

Answer: (c) c = 0

(0, 0) lies on the circle when 0 + 0 + 0 + 0 + c = 0, i.e. c = 0 (as in x² + y² − ax − by = 0). g = 0 puts the centre on the y-axis, g = f puts it on the line y = x, and g² + f² = c makes the radius zero.

Answer: (a) −3 < k < 3

Inside means 1² + k² < 10, i.e. k² < 9, so −3 < k < 3. k = ±3 puts the point on the circle; k < 3 lets k = −5, which is outside.

Answer: (a) 5

Through (0, 0), (6, 0) and (0, −8) the circle is x² + y² − 6x + 8y = 0, with centre (3, −4) and radius √(9 + 16) = 5. (The chord joining (6, 0) and (0, −8) subtends a right angle at O, so it is a diameter of length 10.)

Answer: (c) x² = −6y

x² = −4ay has focus (0, −a). Here x² = −6y gives a = 32 and focus (0, −32). The others have foci (−32, 0), (0, 32) and (32, 0).

Answer: (b) to the left

The equation has a y² term, so the axis is the x-axis; the coefficient of x is negative, so the parabola opens to the left.

Answer: (b) y² = −16x

The latus rectum passes through the focus and is perpendicular to the axis, so the focus is its mid-point (−4, 0) and the axis is the x-axis. Then a = 4 and the parabola opens to the left: y² = −16x. Check: the latus rectum is 16 = 4a.

Answer: (a) (0, −52)

Lying below the x-axis, it is x² = −4ay with 4a = 10, so a = 52 and the focus is (0, −52).