Conic Sections(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

CONIC SECTIONS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

The equation of the circle whose centre is (1, -3) and which touches the line 2x - y - 4 = 0 is

$5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0$
$5 x^{2} + 5 y^{2} + 10 x - 30 y + 49 = 0$
$5 x^{2} + 5 y^{2} - 10 x + 30 y - 49 = 0$
None of these

Discuss Question

Answer: Option A. ->

5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0

:
A
Radius of circle the required circle =

∣ ∣ ∣ \frac{2 + 3 - 4}{\sqrt{5}} ∣ ∣ ∣ = \frac{1}{\sqrt{5}}

Therefore the equation is,

(x - 1)^{2} + (y + 3)^{2} = \frac{1}{5}

i.e,

x^{2} + y^{2} - 2 x + 6 y + 1 + 9 = \frac{1}{5}

i.e,

5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0.

Question 2.

The centres of the circles $x^{2} + y^{2} = 1,$ $x^{2} + y^{2} + 6 x - 2 y = 1$ and $x^{2} + y^{2} - 12 x + 4 y = 1$ are

same
Collinear
Non–collinear
None of these

Discuss Question

Answer: Option B. -> Collinear
:
B
Centre are (0, 0), (-3, 1) and (6, -2) and a line passing through any two points say (0, 0) and (-3, 1) is

y = - \frac{1}{3} x

and point (6, -2) lies on it. Hence points are collinear.

Question 3.

If the line x + 2by + 7 = 0 is diameter of the circle $x^{2} + y^{2} - 6 x + 2 y = 0$ , then b =

3
-5
-1
5

Discuss Question

Answer: Option D. -> 5
:
D
Here the centre of circle (3, -1) must lie on the line x + 2by + 7 = 0
Therefore, 3 - 2b + 7 = 0

\Rightarrow

b = 5

Question 4.

If a straight line through $C (- \sqrt{8}, \sqrt{8})$ making an angle $135^{\circ}$ with the x-axis cuts the circle $x = 5 c o s θ, y = 5 s i n θ$ in points A and B, then length of segment AB is

5
10
15
$15 \sqrt{2}$

Discuss Question

Answer: Option B. -> 10
:
B
Inclination of the line

\leftarrow \to A B

135^{\circ} \Rightarrow S l o p e = t a n 135^{\circ} = - 1

Equation of

\leftarrow \to A B

y - \sqrt{8} = - 1 (x + \sqrt{8}) \Rightarrow x + y = 0

x + y = 0 passes through the centre of the circle

x^{2} + y^{2} = 25

∴

Length of the chord AB = Diameter of the circle =

2 \times 5 = 10

Question 5.

The equation of the circle passing through the point (1, –2) and having its centre on the line 2x – y – 14 = 0 and touching the line 4x + 3y – 23 = 0

$x^{2}$ + $y^{2}$ + 8x + 12y + 27 = 0
$x^{2}$ + $y^{2}$ - 12y + 27 = 0
$x^{2}$ + $y^{2}$ - 8x + 12y - 27 = 0
$x^{2}$ + $y^{2}$ - 8x + 12y + 27 = 0

Discuss Question

Answer: Option D. ->

x^{2}

y^{2}

- 8x + 12y + 27 = 0
:
D
Let P(1, -2) and the centre be C(a, b)
Centre lies on 2x - y - 14 = 0

\Rightarrow

2a - b - 14 = 0

\Rightarrow

b = 2a - 14

\to

(1)
Circle touches the lines 4x + 3y - 23 = 0. Lets say that this point of touching is Q.
We know that,
CQ = CP = Radius of the circle.

\Rightarrow

\sqrt{{(a - 1)}^{2} + {(b + 2)}^{2}}

\frac{| 4 a + 3 b - 23 |}{\sqrt{4^{2} + 3^{2}}}

\Rightarrow

\sqrt{{(a - 1)}^{2} + {(2 a - 14 + 2)}^{2}}

\frac{| 4 a + 3 (2 a - 14) - 23 |}{5}

[From (1)]

\Rightarrow

\sqrt{a^{2} - 2 a + 1 + {4 a}^{2} - 48 a + 144}

| 10 a - 65 |

\Rightarrow

\sqrt{{5 a}^{2} - 50 a + 145}

= 5

| 2 a - 13 |

\Rightarrow

^{2}

- 50a + 145 = (2a - 13)

^{2}

= 4a

^{2}

- 52a + 169

\Rightarrow

^{2}

+ 2a - 24 = 0

\Rightarrow

(a + 6)(a - 4)

\Rightarrow

a = 4
Case 1 : a = 4

\Rightarrow

b = 2(4) - 14 = - 6.

∴

Centre C = (4, - 6)
Radius = CP =

\sqrt{{(4 - 1)}^{2} + {(- 6 + 2)}^{2}}

\sqrt{9 + 16}

= 5
The circle equation is

(x - 4)^{2} + (y + 6)^{2} = 5^{2}

\Rightarrow

x^{2}

y^{2}

- 8x + 12y + 27 = 0
Case 2 : a = - 6

\Rightarrow

b = 2(- 6) - 14 = 26

∴

Centre C = (-6, - 26)
Radius = CP =

\sqrt{{(- 6 - 1)}^{2} + {(- 26 + 2)}^{2}}

\sqrt{49 + 576}

\sqrt{625}

= 25
The circle equation is

(x + 6)^{2} + (y + 26)^{2} = 25^{2}

\Rightarrow

x^{2}

y^{2}

+ 12x + 52y + 87 = 0

∴

The circles are

x^{2}

y^{2}

- 8x + 12y + 27 = 0,

x^{2}

y^{2}

+ 12x + 52y + 87 = 0

Question 6.

If a point P(x,y) moves along the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and if C is the centre of the ellipse, then, 4 max{CP} + 5 min{CP}=___ .

Discuss Question

Answer: Option C. -> 40
:
C

Given ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$
Here, a =5 and b =4
Max{CP} is the maximum distance from the centre to a point on the ellipse. It is equal to the length of the semi-major axis (a).
Similarly, Min {CP} is equal to the length of the semi-minor axis (b).

4 max{CP} + 5 min{CP}= 4a+5b
$= 4 \times 5 + 5 \times 4 = 40$

Question 7.

The line segment joining (5,0) and (10 cos t, 10 sin t) is divided internally in the ratio 2:3 at P. If t varies, then the locus of P is

a straight line
a circle
an ellipse
a hyperbola

Discuss Question

Answer: Option B. -> a circle
:
B

P(x,y) divides the line segment joining (5,0) and (10 cos t, 10 sin t) internally in the ratio 2:3
$∴ x = \frac{2 \times 10 c o s t + 3 \times 5}{2 + 3} x = 4 c o s t + 3 y = \frac{2 \times 10 s i n t + 3 \times 0}{2 + 3} y = 4 s i n t$

$(x - 3)^{2} + y^{2} = 16 c o s^{2} t + 16 s i n^{2} t \Rightarrow (x - 3)^{2} + y^{2} = 16$

The above equation represents a circle

Question 8.

The eccentricity of the hyperbola $4 x^{2} - y^{2} - 8 x - 8 y - 28$ is equal to ___ .

2
$\sqrt{5}$
2
$\sqrt{7}$

Discuss Question

Answer: Option B. ->

\sqrt{5}

:
B

$4 x^{2} - y^{2} - 8 x - 8 y - 28 14 (x - 1)^{2} - 4} - 1 (y + 4)^{2} - 16} - 28 = 0 4 (x - 1)^{2} - (y + 4)^{2} = 16 \frac{(x - 1)^{2}}{4} - \frac{(y + 4)^{2}}{16} = 1$

For the above hyperbola, the eccentricity is given by
$e = \frac{\sqrt{a^{2} + b^{2}}}{a} = \frac{\sqrt{4 + 16}}{2} = \frac{2 \sqrt{5}}{2} = \sqrt{5}$

Question 9.

Given standard equation of ellipse, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b$ , with eccentricity e

Match the following
$\begin{matrix} a)Focus & i) (ae,0) b)Directrix & ii) (a,0) c)Eccentricity & iii) x= \frac{a}{e} d)Vertices & iv) (-ae,0) v) x= \frac{-a}{e} vi) \sqrt{1- \frac{b^{2}}{a^{2}}} \end{matrix}$

a = i, iv, b = ii, v ,c =vi, d =iii
a = i,b = ii, v ,c =vi, d =iii
a = i, iv, b = iii, v ,c =vi, d = ii
a = i,b = iii, v ,c =vi, d = ii

Discuss Question

Answer: Option C. -> a = i, iv, b = iii, v ,c =vi, d = ii
:
C

Let's deduce what all we can infer from $\Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\to f o c u s = (a e, 0), (- a e, 0) \to d i r e c t r i x \Rightarrow x = \frac{a}{e}, x = \frac{- a}{e} \to e c c e n t r i c i t y \Rightarrow \sqrt{1 - \frac{b^{2}}{a^{2}}} \to m a j o r A x i s \Rightarrow y = 0 a s a > b \to m i n o r A x i s \Rightarrow x = 0 a s b < a \to V e r t i c e s \Rightarrow (a, 0), (- a, 0) T h e r e f o r e, a \to i, i v b \to i i i, v c \to v i d \to i i$

Question 10.

If the eccentricity of a hyperbola is $\sqrt{2}$ and if the distance between the foci is 16, then its equation is

$x^{2} - y^{2} = 4$
$x^{2} - y^{2} = 8$
$x^{2} - y^{2} = 16$
$x^{2} - y^{2} = 32$

Discuss Question

Answer: Option D. ->

x^{2} - y^{2} = 32

:
D

Given that distance between foci, 2c =16
$c = 8; e = \sqrt{2}$
$a = \frac{c}{e} = \frac{8}{\sqrt{2}} = 4 \sqrt{2} b = \sqrt{c^{2} - a^{2}} = \sqrt{32} = 4 \sqrt{2}$