**MCQ Questions for Class 11 Maths Chapter 11 Conic Sections 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 = 0Answer

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: (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 = 1Answer

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/9Answer

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 2Answer

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) 1/2Answer

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 = 3²Answer

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 = 0Answer

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: (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 + 8Answer

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) √87/4Answer

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 = b²Answer

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 = 1Answer

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 mAnswer

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)