# MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers

Students are advised to solve the Conic Sections Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Conic Sections Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Conic Sections Class 11 with answers provided with detailed solutions by looking below.

Question 1.

The straight line y = mx + c cuts the circle x² + y² = a² in real points if

(a) √{a² × (1 + m²)} < c

(b) √{a² × (1 – m²)} < c

(c) √{a² × (1 + m²)} > c

(d) √{a² × (1 – m²)} > c

## Answer

Answer: (c) √{a² × (1 + m²)} > c

The straight line y = mx + c cuts the circle x² + y² = a² in real points if

√{a² × (1 + m²)} > c

Question 2.

Equation of the directrix of the parabola x² = 4ay is

(a) x = -a

(b) x = a

(c) y = -a

(d) y = a

## Answer

Answer: (c) y = -a

Given, parabola x² = 4ay

Now, its equation of directrix = y = -a

Question 3.

The equation of parabola with vertex at origin and directrix x – 2 = 0 is

(a) y² = -4x

(b) y² = 4x

(c) y² = -8x

(d) y² = 8x

## Answer

Answer: (c) y² = -8x

Since the line passing through the focus and perpendicular to the directrix is x-axis,

therefore axis of the required parabola is x-axis.

Let the coordinate of the focus is S(a, 0).

Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.

So, 0 = {a – (-2)}/2

⇒ 0 = (a + 2)/2

⇒ a + 2 = 0

⇒ a = -2

Thus the coordinate of focus is (-2, 0)

Let P(x, y) be a point on the parabola.

Then by definition of parabola

(x + 2)² + (y – 0)² = (x – 2)²

⇒ x² + 4 + 4x + y² = x² + 4 – 4x

⇒ 4x + y² = – 4x

⇒ y² = -4x – 4x

⇒ y² = -8x

This is the required equation of the parabola.

Question 4.

The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0

(a) 7

(b) 8

(c) 9

(d) 10

## Answer

Answer: (a) 7

The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)

= {9 + 16 + 10}/√(9 + 16)

= 35/√25

= 35/5

= 7

Question 5.

The equation of a hyperbola with foci on the x-axis is

(a) x²/a² + y²/b² = 1

(b) x²/a² – y²/b² = 1

(c) x² + y² = (a² + b²)

(d) x² – y² = (a² + b²)

## Answer

Answer: (b) x²/a² – y²/b² = 1

The equation of a hyperbola with foci on the x-axis is defined as

x²/a² – y²/b² = 1

Question 6.

If the line 2x – y + λ = 0 is a diameter of the circle x² + y² + 6x − 6y + 5 = 0 then λ =

(a) 5

(b) 7

(c) 9

(d) 11

## Answer

Answer: (c) 9

Given equation of the circle is

x² + y² + 6x − 6y + 5 = 0

Center O = (-3, 3)

radius r = √{(-3)² + (3)² – 5} = √{9 + 9 – 5} = √13

Since diameter of the circle passes through the center of the circle.

So (-3, 3) satisfies the equation 2x – y + λ = 0

⇒ -3 × 2 – 3 + λ = 0

⇒ -6 – 3 + λ = 0

⇒ -9 + λ = 0

⇒ λ = 9

Question 7.

The number of tangents that can be drawn from (1, 2) to x² + y² = 5 is

(a) 0

(b) 1

(c) 2

(d) More than 2

## Answer

Answer: (b) 1

Given point (1, 2) and equation of circle is x² + y² = 5

Now, x² + y² – 5 = 0

Put (1, 2) in this equation, we get

1² + 2² – 5 = 1 + 4 – 5 = 5 – 5 = 0

So, the point (1, 2) lies on the circle.

Hence, only one tangent can be drawn.

Question 8.

The equation of the circle x² + y² + 2gx + 2fy + c = 0 will represent a real circle if

(a) g² + f² – c < 0

(b) g² + f² – c ≥ 0

(c) always

(d) None of these

## Answer

Answer: (b) g² + f² – c ≥ 0

Given, equation of the circle is: x² + y² + 2gx + 2fy + c = 0

This equation can be written as

{x – (-g)}² + {y – (-f)}² + = √{g² + f² – c}²

So, the circle is real is g² + f² – c ≥ 0

Question 9.

The equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is

(a) 16x² – 9y² – 24xy – 144x + 8y + 224 = 0

(b) 16x² + 9y² – 24xy – 144x + 8y – 224 = 0

(c) 16x² + 9y² – 24xy – 144x – 8y + 224 = 0

(d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

## Answer

Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

Given focus S(3, 0)

and equation of directrix is: 3x + 4y = 1

⇒ 3x + 4y – 1 = 0

Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix

Then, SP = PM

⇒ SP² = PM²

⇒ (x – 3)² + (y – 0)² = {(3x + 4y – 1) /{√(3² + 4²)}²

⇒ x² + 9 – 6x + y² = (9x² + 16y² + 1 + 24xy – 8y – 6x)/25

⇒ 25(x² + 9 – 6x + y²) = 9x² + 16y² + 1 + 24xy – 8y – 6x

⇒ 25x² + 225 – 150x + 25y² = 9x² + 16y² + 1 + 24xy – 8y – 6x

⇒ 25x² + 225 – 150x + 25y² – 9x² – 16y² – 1 – 24xy + 8y + 6x = 0

⇒ 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

This is the required equation of parabola.

Question 10.

If the parabola y² = 4ax passes through the point (3, 2), then the length of its latusrectum is

(a) 2/3

(b) 4/3

(c) 1/3

(d) 4

## Answer

Answer: (b) 4/3

Since, the parabola y² = 4ax passes through the point (3, 2)

⇒ 2² = 4a × 3

⇒ 4 = 12a

⇒ a = 4/12

⇒ a = 1/3

So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Question 11.

The eccentricity of an ellipse is?

(a) e = 1

(b) e < 1

(c) e > 1

(d) 0 < e < 1

## Answer

Answer: (d) 0 < e < 1

The eccentricity of an ellipse e = √(1 – a²/b²) and 0 < e < 1

Question 12.

If the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is

(a) (x + 2)² + (y – 3)² = 3²

(b) (x – 2)² + (y + 3)² = 3²

(c) (x – 2)² + (y – 3)² = 3²

(d) (x + 2)² + (y + 3)² = 3²

## Answer

Answer: (c) (x – 2)² + (y – 3)² = 3²

Radius of the circle = √{(2 – 0)² + (3 – 0)² – 2²}

= √(4 + 9 – 4)

= √9

= 3

So, the equation of the circle = (x – 2)² + (y – 3)² = 3²

Question 13.

If the length of the major axis of an ellipse is three times the length of the minor axis then its eccentricity is

(a) 1/3

(b) 1/√3

(c) 1/√2

(d) 2√2/√3

## Answer

Answer: (d) 2√2/√3

Given, the length of the major axis of an ellipse is three times the length of the minor axis

⇒ 2a = 3(2b)

⇒ 2a = 6b

⇒ a = 3b

⇒ a² = 9b²

⇒ a² = 9a² (1 – e²) {since b² = a²(1 – e²)}

⇒ 1 = 9(1 – e²)

⇒ 1/9 = 1 – e²

⇒ e² = 1 – 1/9

⇒ e² = 8/9

⇒ e = √(8/9)

⇒ e = 2√2/√3

So, the eccentricity of the ellipse is 2√2/√3

Question 14.

The equation of parabola with vertex at origin and directrix x – 2 = 0 is

(a) y² = -4x

(b) y² = 4x

(c) y² = -8x

(d) y² = 8x

## Answer

Answer: (c) y² = -8x

Since the line passing through the focus and perpendicular to the directrix is x-axis,

therefore axis of the required parabola is x-axis.

Let the coordinate of the focus is S(a, 0).

Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.

So, 0 = {a – (-2)}/2

⇒ 0 = (a + 2)/2

⇒ a + 2 = 0

⇒ a = -2

Thus the coordinate of focus is (-2, 0)

Let P(x, y) be a point on the parabola.

Then by definition of parabola

(x + 2)² + (y – 0)² = (x – 2)²

⇒ x² + 4 + 4x + y² = x² + 4 – 4x

⇒ 4x + y² = – 4x

⇒ y² = -4x – 4x

⇒ y² = -8x

This is the required equation of the parabola.

Question 15.

In an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is

(a) 4/5

(b) 1/√52

(c) 3/5

(d) 1/2

## Answer

Answer: (c) 3/5

Given, distance between foci = 6

⇒ 2ae = 6

⇒ ae = 3

Again minor axis = 8

⇒ 2b = 8

⇒ b = 4

⇒ b² = 16

⇒ a² (1 – e²) = 16

⇒ a² – a² e² = 16

⇒ a² – (ae)² = 16

⇒ a² – 3² = 16

⇒ a² – 9 = 16

⇒ a² = 9 + 16

⇒ a² = 25

⇒ a = 5

Now, ae = 3

⇒ 5e = 3

⇒ e = 3/5

So, the eccentricity is 3/5

Question 16.

One of the diameters of the circle x² + y² – 12x + 4y + 6 = 0 is given by

(a) x + y = 0

(b) x + 3y = 0

(c) x = y

(d) 3x + 2y = 0

## Answer

Answer: (b) x + 3y = 0

The coordinate of the centre of the circle x² + y² – 12x + 4y + 6 = 0 are (6, -2)

Clearly, the line x + 3y passes through this point.

Hence, x + 3y = 0 is a diameter of the given circle.

Question 17.

The center of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is?

(a) (2, -3)

(b) (-2, 3)

(c) (-4, 6)

(d) (4, -6)

## Answer

Answer: (a) (2, -3)

Given, equation of the circle is 4x² + 4y² – 8x + 12y – 25 = 0

⇒ x² + y² – 8x/4 + 12y/4 – 25/4 = 0

⇒ x² + y² – 2x + 3y – 25/4 = 0

Now, center = {-(-2), -3} = (2, -3)

Question 18.

If the parabola y² = 4ax passes through the point (3, 2), then the length of its latusrectum is

(a) 2/3

(b) 4/3

(c) 1/3

(d) 4

## Answer

Answer: (b) 4/3

Since, the parabola y² = 4ax passes through the point (3, 2)

⇒ 2² = 4a × 3

⇒ 4 = 12a

⇒ a = 4/12

⇒ a = 1/3

So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Question 19.

The equation of ellipse whose one focus is at (4, 0) and whose eccentricity is 4/5 is

(a) x²/5 + y²/9 = 1

(b) x2 /25 + y² /9 = 1

(c) x²/9 + y²/5 = 1

(d) x²/9 + y²/25 = 1

## Answer

Answer: (b) x² /25 + y² /9 = 1

Given focus is (4, 0)

⇒ ae = 4

and e = 4/5

a × (4/5) = 4

⇒ a = 5

Now, b² = a² (1 – e²)

⇒ b² = 5² {1 – (4/5)²}

⇒ b² = 25{1 – 16/25}

⇒ b² = 25{(25 – 16)/25}

⇒ b² = 9

Hence, the equation of the ellipse is x²/a² + y²/b² = 1

⇒ x²/5² + y²/9 = 1

⇒ x²/25 + y²/9 = 1

Question 20.

The focus of parabola y² = 8x is

(a) (2, 0)

(b) (-2, 0)

(c) (0, 2)

(d) (0, -2)

## Answer

Answer: (a) (2, 0)

Given, y² = 8x

General equation is y² = 4ax

Now, 4a = 8

⇒ a = 2

Now, focus = (a, 0) = (2, 0)

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