Class 11 Maths Chapter 10 Conic Sections Worksheet with Answers, free PDF (CBSE Class 11 and JEE Main; based on NCERT)

A 20-mark worksheet on Conic Sections for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

Which of the following parabolas has its focus on the negative y-axis?

(a) y² = −6x(b) x² = 6y(c) x² = −6y(d) y² = 6x

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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).

Question 2

The equation x² + y² + 2gx + 2fy + c = 0 represents a circle that passes through the origin exactly when

(a) g = 0(b) g = f(c) c = 0(d) g² + f² = c

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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.

Question 3

The point (1, k) lies inside the circle x² + y² = 10 exactly when

(a) −3 < k < 3(b) k < 3(c) k = ±3(d) −9 < k < 9

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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.

Question 4

The latus rectum of a parabola with vertex at the origin has end points (−4, 8) and (−4, −8). The equation of the parabola is

(a) y² = 16x(b) y² = −16x(c) x² = −16y(d) y² = −8x

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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.

Question 5

A plane passes through the vertex of a double-napped cone and contains one of its generators, so that β = α. The section is

(a) a point(b) a pair of intersecting straight lines(c) a circle(d) a straight line

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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.

Question 6 · 2 marks

Find the points of the parabola x² = 8y whose distance from the focus is 10 units.

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4a = 8, so a = 2: the focus is (0, 2) and the directrix y = −2. By the definition, the distance from the focus equals the distance from the directrix, y + 2 = 10, so y = 8. Then x² = 64 and x = ±8: the points are (8, 8) and (−8, 8).

focus (0, 2), directrix y = −2 and the equation y + 2 = 10 (1)
y = 8 and the points (±8, 8) (1)

Question 7 · 2 marks

The eccentricity of the ellipse x²a² + y²b² = 1 (a > b) is 12 and its minor axis is 6√3 units long. Find a and write the equation of the ellipse.

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b = 3√3, so b² = 27. b² = a² − c² = a²(1 − e²) = 34a², so a² = 36 and a = 6. The ellipse is x²36 + y²27 = 1.

b² = a²(1 − e²) with b² = 27 (1)
a = 6 and x²36 + y²27 = 1 (1)

Question 8 · 2 marks

Show that the equation x² + y² + 4x − 6y + 15 = 0 does not represent a circle.

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Completing the squares: (x² + 4x + 4) + (y² − 6y + 9) = −15 + 4 + 9, i.e. (x + 2)² + (y − 3)² = −2. A sum of two squares cannot be negative, so no point (x, y) satisfies the equation, and it is not a circle (g² + f² − c = 4 + 9 − 15 = −2 < 0).

completes the squares to (x + 2)² + (y − 3)² = −2 (1)
concludes no real point satisfies it, so it is not a circle (1)

Question 9 · 3 marks

A point P moves so that the difference of its distances from A(0, −3) and B(0, 3) is always 4. Show, from this condition, that the locus of P is a hyperbola, and write its equation in standard form.

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√(x² + (y + 3)²) − √(x² + (y − 3)²) = ±4. Then √(x² + (y + 3)²) = ±4 + √(x² + (y − 3)²); squaring, 12y − 16 = ±8√(x² + (y − 3)²), i.e. 3y − 4 = ±2√(x² + (y − 3)²). Squaring again, 9y² − 24y + 16 = 4x² + 4y² − 24y + 36, so 5y² − 4x² = 20, i.e. y²4 − x²5 = 1: a hyperbola with foci (0, ±3), a = 2 and b² = 9 − 4 = 5.

writes the distance condition and isolates one root (1)
squares twice correctly (1)
y²4 − x²5 = 1 (1)

Question 10 · 3 marks

A(−1, 2) and B(5, −6) are the ends of a diameter of a circle. Find the equation of the circle, and decide whether the point (6, 1) lies inside, on or outside it.

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The centre is the mid-point (2, −2) of AB, and AB = √(36 + 64) = 10, so r = 5. The circle is (x − 2)² + (y + 2)² = 25, i.e. x² + y² − 4x + 4y − 17 = 0. For (6, 1): (6 − 2)² + (1 + 2)² = 16 + 9 = 25 = r², so the point lies on the circle.

centre (2, −2) and radius 5 (1)
(x − 2)² + (y + 2)² = 25 (1)
(6 − 2)² + (1 + 2)² = 25: on the circle (1)

Question 11 · 3 marks

The triangle formed by the vertex of the parabola y² = 4ax (a > 0) and the two ends of its latus rectum has area 50 square units. Find the equation of the parabola and of its directrix.

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The ends of the latus rectum are (a, 2a) and (a, −2a): base 4a, at a distance a from the vertex O. Area = 12 × 4a × a = 2a² = 50, so a = 5. The parabola is y² = 20x and the directrix x = −5.

ends (a, ±2a) and area = 2a² (1)
a = 5 (1)
y² = 20x and directrix x = −5 (1)

Class 11 Maths Chapter 10: Conic Sections worksheet with answer key