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
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
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) 2/5Answer: (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
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
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
Question 5
Find the coordinates of foci of hyperbola
a) (±5,0)
b) (±4,0)
c) (0,±5)
d) (0,±4)
Answer c
Question 6
What is transverse axis length for hyperbola
a) 5 units
b) 4 units
c) 8 units
d) 6 units
Answer d
Question 7
What is conjugate axis length for hyperbola
a) 5 units
b) 4 units
c) 8 units
d) 10 units
Answer c
Question 8
What is equation of latus rectums of hyperbola
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
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