Maths MCQ Class XI Ch-11 | Conic Section

 Mathematics MCQ Class XI Chapter-11

Conic Section

MCQ BASED ON CIRCLE CLASS XI

Question 1

Find the equation of circle with centre at origin and radius
5 units.
a) x² + y² = 25
b) x² + y² = 5
c) x² = 25
d) y² = 25

Answer a

Question 2

The point (6, 2) lie ___________ the circle x² + y² – 2x -4y – 36 = 0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Answer c

Question 3

Find the equation of circle with centre at (2, 5) and radius
5 units.
a) x² + y²   + 4x – 10y + 4 = 0
b) x² + y²  – 4x – 10y +
4 = 0
c) x² + y²  + 4x + 10y +
4 = 0
d) x² + y²  + 4x – 10y –
4 = 0

Answer b

Question 4

Find the centre of the circle with equation x² + y² – 4x – 10y + 4 =
0.
a) (-2, 5)
b) (-2, -5)
c) (2, -5)
d) (2, 5)

Answer: d
Explanation: Comparing the equation with general
form x² + y² + 2gx + 2fy + c =
0, we get
2g = – 4 ⇒ g = – 2
2f = -10 ⇒ f = – 5
c = 4
Centre is at (-g, -f) i.e. (2, 5).

Question 5

Find the radius of the circle with equation x² + y² – 4x – 10y + 4 =
0.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c

Question 6

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

a) (-2, 3)

b) (1, -3/2)

c) (-4, 6)

d) (4, -6)

Answer: (b)
(1, -3/2)

Explanation:

Given circle
equation: 4x² + 4y² – 8x + 12y – 25 = 0

⇒ x² + y² –
(8x/4) + (12y/4) – (25/4) = 0

⇒ x² +y² -2x
+3y -(25/4) = 0 …(1)

As we know that the
general equation of a circle is x² + y² + 2gx + 2fy + c = 0,
and the centre of the circle = (-g, -f)

Hence, by comparing
equation (1) and the general equation,

2g = -2,  ⇒ g = -1

2f = 3,    ⇒ f = 3/2

Now, substitute the
values in the centre of the circle (-g, -f), we get,

Centre = (1, -3/2).

Question 7

If a circle pass through (2, 0) and (0, 4) and centre at
x-axis then find the radius of the circle.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c
Explanation: Equation of circle with centre at
x-axis (a, 0) and radius r units is
(x – a)² + (y)² = r²
⇒ (2 – a)² + (0)² = r²
And (0 – a)² + (4)² = r²
⇒ (a – 2)² = a² + 4² ⇒ (- 2)(2a
– 2) =16 ⇒ a – 1 = – 4 ⇒ a = – 3

So, r² = (2 + 3)² = 5 ²  ⇒ r = 5 units.

Question 8
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 a

Question 9

If a circle pass through (4, 0) and (0, 2) and centre at
y-axis then find the radius of the circle.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c

Question 10

The point (1, 4) lie ___________ the circle x² + y² – 2x – 4y + 2 = 0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Answer b

Question 11
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 c

MCQ BASED ON PARABOLA CLASS XI

Question 1

Find the focus of parabola with equation y² = 100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer c

Question 2

Find the focus of parabola with equation y² = – 100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer d

Question 3

Find the focus of parabola with equation x² = 100y.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer a

Question 4

The focus of the
parabola y² = 8x is

a) (0,
2)

b) (2,
0)

c) (0,
-2)

d) (-2,
0)

Answer: (b)
(2, 0)

Explanation:

Given parabola
equation y² = 8x …(1)

Here, the
coefficient of x is positive and the standard form of parabola is y² =
4ax …(2)

Comparing (1) and
(2), we get

4a = 8

a = 8/4 = 2

We know that the
focus of parabolic equation y² = 4ax is (a, 0).

Therefore, the
focus of the parabola y² =8x is (2, 0).

Question 5

The length of
the latus rectum of x² = -9y is equal to

a)  3 units

b) – 3 units

c)  9/4 units

d)  9 units

Answer: (d) 9
units

Explanation:

Given parabola
equation: x² = – 9y …(1)

Since the
coefficient of y is negative, the parabola opens downwards.

The general
equation of parabola is x² = – 4ay…(2)

Comparing (1) and
(2), we get

-4a = -9

a = 9/4

We know that the
length of latus rectum = 4a = 4(9/4) = 9.

Therefore, the
length of the latus rectum of x² = -9y is equal to 9 units.

Question 6

The parametric
equation of the parabola y² = 4ax is

a) x = at; y = 2at

b) x = at²;
y = 2at

c) x = at²;
y² = at³

d) x = at²;
y = 4at

Answer: (b) x
= at²; y = 2at

Question 7

Find the equation of latus rectum of parabola y² = 100x.

a) x = 25
b) x = – 25
c) y = 25
d) y = – 25

Answer a

Question 8

Find the equation of latus rectum of parabola x² = – 100y.
a) x = 25
b) x = – 25
c) y = – 25
d) y = 25

Answer c

Question 9

Find the equation of directrix of parabola y²=-100x.
a) x = 25
b) x = – 25
c) y = – 25
d) y = 25

Answer a

Question 10
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 c

Question 11

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
(d) y² = 2x/9

Answer b

Question 12

Find the vertex of the parabola y² =
4ax.
a) (0, 4)
b) (0, 0)
c) (4, 0)
d) (0, -4)

Answer b

Question 13

Find the equation of axis of the parabola y² = 24x.
a) x = 0
b) x = 6
c) y = 6
d) y = 0

Answer d

Question 14

Find the equation of axis of the parabola x² = 24y.
a) x = 0
b) x = 6
c) y = 6
d) y = 0

Answer a

Question 15

Find the length of latus rectum of the parabola y² = 40x.

a) 4 units
b) 10 units
c) 40 units
d) 80 units

Answer c

Question 16

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 b

MCQ BASED ON ELLIPSE CLASS – XI

Question 1

An ellipse has ___________ vertices and ____________ foci.
a) two, one
b) one, one
c) one, two
d) two, two

Answer d

Question 2

Find the coordinates of foci of ellipse (x/25)² + (y/16)² = 1.
a) (±3, 0)
b) (±4, 0)
c) (0, ±3)
d) (0, ±4)

Answer a

Question 3
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 c

Question 4
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 a

Question 5

Find the coordinates of foci of ellipse (x/16)² + (y/25)² = 1.
a) (±3, 0)
b) (±4, 0)
c) (0, ±3)
d) (0, ±4)

Answer c

Question 6

What is major axis length for ellipse (x/25)² + (y/16)² = 1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units

Answer d

Question 7

For the ellipse
3x² + 4y² = 12, the length of the latus rectum is:

a)  2/5
b)  3/5
c)  3
d) 4

Answer: (c) 3

Explanation:

Given ellipse
equation: 3x² + 4y² = 12

The given equation
can be written as (x²/4) + (y²/3) = 1…(1)

Now, compare the
given equation with the standard ellipse equation: (x²/a²)
+ (y²/b²) = 1, we get

a = 2 and b = √3

Therefore, a >
b.

If a>b, then the
length of latus rectum is 2b²/a

Substituting the
values in the formula, we get

Length of latus
rectum = [2(√3)²] /2 = 3

Question 8

What is minor axis length for ellipse (x/25)² + (y/16)² = 1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units

Answer c

Question 9

What is equation of latus rectums of ellipse (x/25)² + (y/16)² = 1?

a) x = ±3
b) y = ±3
c) x = ±2
d) y = ±2

Answer a

Question 10

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

a) 1/2

b)  1/5

c)   3/5

d)   4/5

Answer: (c)
3/5

Explanation:

Given that the
minor axis of ellipse is 8.(i. e) 2b = 8. So, b=4.

Also, the distance
between its foci is 6. (i. e) 2ae = 6

Therefore, ae = 6/2
= 3

We know that b² =
a²(1-e²)

b² =
a² – a²e²

b² =
a² – (ae)²

Now, substitute the
values to find the value of a.

(4)² =
a² -(3)²

16 = a² –
9

a² =
16+9 = 25.

So, a = 5.

The formula to
calculate the eccentricity of ellipse is e = √[1-(b²/a²)]

e = √[1-(4²/5²)]

e = √[(25-16)/25]

e = √(9/25) = 3/5.

Question 11

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 d

MCQ BASED ON HYPERBOLA CLASS XI

Question 1

A hyperbola has ___________ vertices and ____________ foci.

a) two, one

b) one, one

c) one, two

d) two, two

Answer d

Question 2

Find the coordinates of foci of hyperbola (x/9)²−(y/16)²=1.

a) (±5,0)

b) (±4,0)

c) (0,±5)

d) (0,±4)

Answer a

Question 3
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 b

The eccentricity of hyperbola is
a. e =1
b. e > 1
c. e < 1
d. 0 < e < 1
Answer: (b) e > 1

Question 5

Find the coordinates of foci of hyperbola (y/16)²−(/x9)²=1.

a) (±5,0)

b) (±4,0)

c) (0,±5)

d) (0,±4)

Answer c

Question 6

What is transverse axis length for hyperbola (x/9)²−(y/16)²=1?

a) 5 units

b) 4 units

c) 8 units

d) 6 units

Answer d

Question 7

What is conjugate axis length for hyperbola (x/9)²−(y/16)²=1?

a) 5 units

b) 4 units

c) 8 units

d) 10 units

Answer c

Question 8

What is equation of latus rectums of hyperbola (x/9)²−(y/16)²=1?

a) x = ±5

b) y = ±5

c) x = ±2

d) y = ±2

Answer a 

Question 9

The length of
the transverse axis is the distance between the ____.

a)   Two vertices

b)   Two Foci

c)   Vertex and the
origin

d)   Focus and the
vertex

Answer: (a)
Two vertices