Check the below NCERT MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Conic Sections Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

## Conic Sections Class 11 MCQs Questions with Answers

Question 1.

The locus of the point from which the tangent to the circles x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0 are equal is given by the equation

(a) 8x + 19 = 0

(b) 8x – 19 = 0

(c) 4x – 19 = 0

(d) 4x + 19 = 0

## Answer

Answer: (b) 8x – 19 = 0

Hint:

Given equation of circles are x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0

Now, the required line is the radical axis of the two circles are

(x² + y² – 4) – (x² + y² – 8x + 15) = 0

⇒ x² + y² – 4 – x² – y² + 8x – 15 = 0

⇒ 8x – 19 = 0

Question 2.

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

Hint:

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

= {9 + 16 + 10}/√(9 + 16)

= 35/√25

= 35/5

= 7

Question 3.

A man running a race course notes that the sum of the distances from the two flag posts from him is always 10 meter and the distance between the flag posts is 8 meter. The equation of posts traced by the man is

(a) x²/9 + y²/5 = 1

(b) x²/9 + y2 /25 = 1

(c) x²/5 + y²/9 = 1

(d) x²/25 + y²/9 = 1

## Answer

Answer: (d) x²/25 + y²/9 = 1

Hint:

From the question, it is clear that the path traced by the man is an ellipse having its foci at two posts.

Let the equation of the ellipse be

x²/a² + y²/b² = 1

It is given that the sum of the distances of the man from the two flag posts is 10 m

This means that the sum of focal distances of a point on the ellipse is 10 m

⇒ PS + PS_{1} = 10

⇒ 2a = 10

⇒ a = 5

Again, given that the distance between the flag posts is 8 meters

⇒ 2ae = 8

⇒ ae = 4

Now, b² = a² (1 – e²)

⇒ b² = a² – a² e²

⇒ b² = a² – (ae)²

⇒ b² = 5² – 4²

⇒ b² = 25 – 16

⇒ b² = 9

⇒ b = 3

Hence, the equation of the path is x²/5² + y²/3² = 1

⇒ x²/25 + y²/9 = 1

Question 4.

The center of the ellipse (x + y – 2)² /9 + (x – y)² /16 = 1 is

(a) (0, 0)

(b) (0, 1)

(c) (1, 0)

(d) (1, 1)

## Answer

Answer: (d) (1, 1)

Hint:

The center of the given ellipse is the point of intersection of the lines

x + y – 2 = 0 and x – y = 0

After solving, we get

x = 1, y = 1

So, the center of the ellipse is (1, 1)

Question 5.

The parametric coordinate of any point of the parabola y² = 4ax is

(a) (-at², -2at)

(b) (-at², 2at)

(c) (a sin²t, -2a sin t)

(d) (a sin t, -2a sin t)

## Answer

Answer: (c) (a sin²t, -2a sin t)

Hint:

The point (a sin²t, -2a sin t) satisfies the equation of the parabola y² = 4ax for all

values of t. So, the parametric coordinate of any point of the parabola y² = 4ax is

(a sin²t, -2a sin t)

Question 6.

The equation of parabola with vertex at origin the axis is along x-axis and passing through the point (2, 3) is

(a) y² = 9x

(b) y² = 9x/2

(c) y² = 2x

(b) y² = 2x/9

## Answer

Answer: (b) y² = 9x/2

Hint:

A parabola with its axis along the x-axis and vertex(0, 0) and direction x = -a has the equation:

y² = 4ax ………….. 1

Given, point (2,3) lies on the parabola,

⇒ 3² = 4a × 2

⇒ 9 = 4a × 2

⇒ 9/2 = 4a

From equation 1, we get

y² = (9/2)x

⇒ y² = 9x/2

This is the required equation of the parabola.

Question 7.

At what point of the parabola x² = 9y is the abscissa three times that of ordinate

(a) (1, 1)

(b) (3, 1)

(c) (-3, 1)

(d) (-3, -3)

## Answer

Answer: (b) (3, 1)

Hint:

Given, parabola is x² = 9y

Let P(h, k) is the point on the parabola such that abscissa is 3 times the ordinate.

So, h = 3k ……… 1

Since P(h, k) lies on the parabola

So, h² = 9k ……… 2

From equation 1 and 2, we get

(3k)² = 9k

⇒ 9k² = 9k

⇒ 9k² – 9k = 0

⇒ 9k(k – 1) = 0

⇒ k = 0, 1

When k = 0, h = 0

So k = 1

Now, from equation 1,

h = 3 × 1 = 3

So, the point is (3, 1)

Question 8.

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

Hint:

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 9.

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

Hint:

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 10.

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

Hint:

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 11.

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

Hint:

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 12.

The parametric representation (2 + t², 2t + 1) represents

(a) a parabola

(b) a hyperbola

(c) an ellipse

(d) a circle

## Answer

Answer: (a) a parabola

Hint:

Let x = 2 + t²

⇒ x – 2 = t² ……….. 1

and y = 2t + 1

⇒ y – 1 = 2t

⇒ (y – 1)/2 = t

From equation 1, we get

x – 2 = {(y – 1)/2}²

⇒ x – 2 = (y – 1)²/4

⇒ (y – 1)² = 4(x – 2)

This represents the equation of a parabola.

Question 13.

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

Hint:

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

x²/a² – y²/b² = 1

Question 14.

The equation of parabola with vertex (-2, 1) and focus (-2, 4) is

(a) 10y = x² + 4x + 16

(b) 12y = x² + 4x + 16

(c) 12y = x² + 4x

(d) 12y = x² + 4x + 8

## Answer

Answer: (b) 12y = x² + 4x + 16

Hint:

Given, parabola having vertex is (-2, 1) and focus is (-2, 4)

As the vertex and focus share the same abscissa i.e. -2,

parabola axis of symmetry as x = -2

⇒ x + 2 = 0

Hence, the equation of a parabola is of the type

(y – k) = a(x – h)² where (h, k) is vertex

Now, focus = (h, k + 1/4a)

Since, vertex is (-2, 1) and parabola passes through vertex

So, focus = (-2, 1 + 1/4a)

Now, 1 + 1/4a = 4

⇒ 1/4a = 4 -1

⇒ 1/4a = 3

⇒ 4a = 1/3

⇒ a = /1(3 × 4)

⇒ a = 1/12

Now, equation of parabola is

(y – 1) = (1/12) × (x + 2)²

⇒ 12(y – 1) = (x + 2)²

⇒ 12y – 12 = x² + 4x + 4

⇒ 12y = x² + 4x + 4 + 12

⇒ 12y = x² + 4x + 16

This is the required equation of parabola.

Question 15.

If a parabolic reflector is 20 cm in diameter and 5 cm deep then the focus of parabolic reflector is

(a) (0 0)

(b) (0, 5)

(c) (5, 0)

(d) (5, 5)

## Answer

Answer: (c) (5, 0)

Hint:

given diameter of the parabola is 20 m.

The equation of parabola is y² = 4ax.

Since this parabola passes through the point A(5,10) then

10² = 4a×5

⇒ 20a = 100

⇒ a = 100/20

⇒ a = 5

So focus of parabola is (a, 0) = (5, 0)

Question 16.

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

(a) √57/4

(b) √77/4

(c) √77/2

(d) √87/4

## Answer

Answer: (c) √77/2

Hint:

Given, equation fo the 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, radius = √{(-2)² + (3)² – (-25/4)}

= √{4 + 9 + 25/4}

= √{13 + 25/4}

= √{(13×4 + 25)/4}

= √{(52 + 25)/4}

= √{77/4}

= √77/2

Question 17.

If (a, b) is the mid point of a chord passing through the vertex of the parabola y² = 4x, then

(a) a = 2b

(b) 2a = b

(c) a² = 2b

(d) 2a = b²

## Answer

Answer: (d) 2a = b²

Hint:

Let P(x, y) be the coordinate of the other end of the chord OP where O(0, 0)

Now, (x + 0)/2 = a

⇒ x = 2a

and (y + 0)/2 = b

⇒ y = 2b

Now, y² = 4x

⇒ (2b)² = 4 × 2a

⇒ 4b² = 8a

⇒ b² = 2a

Question 18.

A rod of length 12 CM moves with its and always touching the co-ordinate Axes. Then the equation of the locus of a point P on the road which is 3 cm from the end in contact with the x-axis is

(a) x²/81 + y²/9 = 1

(b) x²/9 + y²/81 = 1

(c) x²/169 + y²/9 = 1

(d) x²/9 + y²/169 = 1

## Answer

Answer: (a) x²/81 + y²/9 = 1

Hint:

Given a rod of length 12 cm moves with its ends always touching the coordinate axes.

Again given a point P on the rod, which is 3 cm from the end in contact with the x-axis.

It is shown in the figure.

Here AP = 3 cm, AB = 12

Now BP = AB – AP

⇒ BP = 12 – 3

⇒ BP = 9 cm

Again from figure,

∠PAO = ∠BPO = θ (since PQ || OA and are corresponding angles)

Now in ΔBPO,

cosθ = QP/BP

⇒ cosθ = x/9 …………. 1

Again in ΔPAr,

sinθ = PR/PA

⇒ sinθ = y/3 …….. 2

Now square equation 1 and 2 and then add them, we get

cos² θ + sin² θ = x²/81 + y²/9

⇒ x²/81 + y²/9 = 1 (since cos² θ + sin² θ = 1 )

So, the equation of the locus of a point P is x²/81 + y²/9 = 1

Question 19.

The line lx + my + n = 0 will touches the parabola y² = 4ax if

(a) ln = am²

(b) ln = am

(c) ln = a² m²

(d) ln = a² m

## Answer

Answer: (a) ln = am²

Hint:

Given, lx + my + n = 0

⇒ my = -lx – n

⇒ y = (-l/m)x + (-n/m)

This will touches the parabola y² = 4ax if

(-n/m) = a/(-l/m)

⇒ (-n/m) = (-am/l)

⇒ n/m = am/l

⇒ ln = am²

Question 20.

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)

Hint:

Given, equation fo the 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)

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Conic Sections MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

## Leave a Reply