Conic Sections Test

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Conic Sections Test - 48

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Question 1
1 / -0

Find the equation of a circle whose centre is (2, - 1 ) an radius is 3

A
$$

x ^ { 2 } + y ^ { 2 } + 4 x - 2 y + 4 = 0

$$

B
$$

x ^ { 2 } + y ^ { 2 } - 4 x + 2 y - 4 = 0

$$

C
$$

x ^ { 2 } + y ^ { 2 } + 4 x + 2 y - 4 = 0

$$

D
$$

x ^ { 2 } + y ^ { 2 } + 2 x - 4 y - 4 = 0

$$

Solution

Center $$=(2,-1)$$ and $$r=3$$ [ Given ]
We know that, center $$=(-g,-f)$$
So, by comparing we get, $$g=-2$$ and $$f=1$$
$$r=\sqrt{g^2+f^2-c}$$
$$\Rightarrow$$ $$3=\sqrt{(-2)^2+(1)^2-c}$$
$$\Rightarrow$$ $$9=4+1-c$$
$$\Rightarrow$$ $$9=5-c$$
$$\therefore$$ $$c=-4$$
The equation of circle is
$$\Rightarrow$$ $$x^2+y^2+2gx+2fy+c=0$$
$$\Rightarrow$$ $$x^2+y^2+2(-2)x+2(1)y-4=0$$ [ Substituting values of $$g,f$$ and $$c$$ ]
$$\Rightarrow$$ $$x^2+y^2-4x+2y-4=0$$
Question 2
1 / -0

In what ratio, the point of intersection of the common tangents to hyperbola $$\dfrac{x^2}{1}-\dfrac{y^2}{8}=1$$ and parabola $$y^2=12x$$, divides the foci of the given hyperbola?

A
$$3:4$$

B
$$3:2$$

C
$$5:4$$

D
$$5:3$$

Solution

Let equation of common tangent is $$y=mx+\dfrac{3}{m}$$
$$\therefore \left(\dfrac{3}{m}\right)^2=1.m^2-8$$
$$\Rightarrow m^4-8m^2-9=0$$
$$\Rightarrow m^2=9$$
$$\Rightarrow m=\pm 3$$
$$\therefore$$ equation of common tangents are $$y=3x+1$$ & $$y=-3x+1$$
$$\therefore \dfrac{PS}{PS'}=\dfrac{3+\dfrac{1}{3}}{-\dfrac{1}{3}+3}=\dfrac{5}{4}$$.
Question 3
1 / -0

$$AB$$ is a chord of the circle $$x^{2} + y^{2} = 9$$. The tangent at $$A$$ and $$B$$ intersect at $$C$$. If $$(1, 2)$$ is the midpoint of $$AB$$, the area of $$\triangle ABC$$ is (in square units).

A
$$9$$

B
$$\dfrac {7}{\sqrt {5}}$$

C
$$\dfrac {9}{\sqrt {5}}$$

D
$$\dfrac {8}{\sqrt {5}}$$

Solution

The equation of chord with midpoint $$(1, 2)$$ in the circle $$x^{2} + y^{2} = 9$$ is
$$x \times 1 + y\times 2 - 9 = 1^{2} + 2^{2} - 9$$
$$\Rightarrow x + 2y - 5 = 0 .... (1)$$
Also, if the point of intersection of tangents $$C$$ has coordinates $$(h, k)$$, then the equation of common chord is
$$hx + ky - 9 = 0 .... (2)$$
Since (1) and (2) represent the same equation, we get
$$\dfrac {h}{1} = \dfrac {k}{2} = \dfrac {9}{5}$$
$$\Rightarrow h = \dfrac {9}{5}, k = \dfrac {18}{5}$$
From figure, $$AD = \sqrt {OA^{2} - OD^{2}} = \sqrt {9 - 5} = 2$$
$$CD = OC - OD = \sqrt {h^{2} + k^{2}} - \sqrt {5} = \dfrac {9}{\sqrt {5}} - \sqrt {5} = \dfrac {4}{\sqrt {5}}$$
The area of $$ABC = \dfrac {1}{2} AB . CD = AD.CD = 2\times \dfrac {4}{\sqrt {5}} = \dfrac {8}{\sqrt {5}}$$.
Question 4
1 / -0

The equation $$\dfrac {x^{2}}{2 - \lambda} + \dfrac {y^{2}}{\lambda - 5} - 1 = 0$$ represents an ellipse, if

A
$$\lambda < 5$$

B
$$\lambda < 2$$

C
$$2 < \lambda < 5$$

D
$$\lambda < 2$$ or $$\lambda < 5$$

Solution

General equation of ellipse is $$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$$

So both denominator should be positive as they are squares

In the given equation

$$\dfrac {x^2}{2-\lambda} +\dfrac {y^2}{\lambda -5}-1=0$$

So,

$$2-\lambda > 0, +(\lambda -5) >o$$

$$\Rightarrow \ \lambda < 2, \lambda > 5$$

$$\Rightarrow \ 2 < \lambda < 5$$
Question 5
1 / -0

The eccentricity of the ellipse $$9x^{2} + 25y^{2} - 18x - 100y - 116 = 0$$, is

A
$$25/16$$

B
$$4/5$$

C
$$16/25$$

D
$$5/4$$

Solution

Given equation of ellipse is

$$9x^2+25y^2-18x-100y-116=0$$

$$9(x^2-2x+1)+25(y^2-4y+4)-225=0$$

$$9(x-1)^2+25(y-2)^2=225$$

$$\dfrac{9(x-1)^2}{225}+\dfrac{25(y-2)^2}{225}=1$$

$$\Rightarrow \dfrac{(x-1)^2}{25}+\dfrac{(y-2)^2}{9}=1$$

comparing with the standard form

$$\dfrac{(x-p)^2}{a^2}+\dfrac{(y-2)^2}{9}=1$$

Here $$a^2>b^2$$

eccentricity $$(e) =\sqrt{\dfrac{a^2-b^2}{a^2}}$$

$$e=\sqrt{\dfrac{25-9}{25}}=\sqrt{\dfrac{16}{25}}$$

$$e=\dfrac45$$
Question 6
1 / -0

The eccentricity of the ellipse $$5x^{2} + 9y^{2} = 1$$ is

A
$$\dfrac 23$$

B
$$\dfrac 34$$

C
$$\dfrac 45$$

D
$$\dfrac 12$$

Solution

Consider the equation of the ellipse is

$$5x^2+9y^2=1$$

$$\dfrac {\dfrac {x^2}{1}}{5}+\dfrac {\dfrac {y^2}{1}}{9}=1$$

Comparing with the standard equation of the ellipse, we get:

$$a^2=1/5 $$ and $$b^2=1/9$$ i.e., $$a=1/ \sqrt 5$$ and $$b=1/3$$

Here $$a > b$$

now, $$e=\sqrt {1-b^2 /a^2}$$

$$\Rightarrow \ e=\sqrt {1-\dfrac {1/9}{1/5}}=\sqrt {1\times 5/9}=\sqrt {4/9}$$

$$e=2/3$$

Hence eccentricity of ellipse $$=2/3$$
Question 7
1 / -0

An ellipse has its centre at $$(1, -1)$$ and semi-major axis $$= 8$$ and it passes through the point $$(1, 3)$$. The equation of the ellipse is

A
$$\dfrac {(x + 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$$

B
$$\dfrac {(x - 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$$

C
$$\dfrac {(x - 1)^{2}}{16} + \dfrac {(y + 1)^{2}}{64} = 1$$

D
$$\dfrac {(x + 1)^{2}}{64} + \dfrac {(y - 1)^{2}}{16} = 1$$

Solution

Given that

centre is at $$(1, -1)$$

semi major axis $$(a)=8$$

so, equation of ellipse can be written as

$$\dfrac {(x-1)^2}{a^2} +\dfrac {(y+1)^2}{b^2} =1.....(1)$$

It passes through point $$(1,3)$$

i.e, $$x=1, y=3$$

Putting these value in equation $$(1)$$ we get

$$\dfrac {(1-1)^2}{a^2} +\dfrac {(3+1)^2}{b^2}=1$$

$$\dfrac {16}{b^2}=1$$

$$b^2=16\ \Rightarrow b=4$$

Substituting the values of $$a$$ and $$b$$ in equation $$(1)$$ we get

$$\dfrac {(x-1)^2}{64}+\dfrac {(y+1)^2}{16}=1$$

This is the required equation of ellipse
Question 8
1 / -0

The eccentricity of the ellipse $$25x^{2} + 16^{2} = 400$$ is

A
$$\dfrac 35$$

B
$$\dfrac 13$$

C
$$\dfrac 25$$

D
$$\dfrac 15$$

Solution

Given equation of ellipse is

$$25x^2+16y^2=400$$

$$\dfrac{25x^2}{400}+\dfrac{16y^2}{400}=1$$

$$\Rightarrow \dfrac{x^2}{16}+\dfrac{y^2}{25}=1$$

comparing with the standard form

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

Here $$b^2>a^2$$

$$\Rightarrow$$ Eccentricity $$(e) =\sqrt{\dfrac{b^2-a^2}{b^2}}$$

$$e=\sqrt{\dfrac{25-16}{25}}=\sqrt{\dfrac{9}{25}}$$

$$e=\dfrac 35$$
Question 9
1 / -0

If the major axis of an ellipse is three times the minor axis, then its eccentricity is equal to

A
$$\dfrac {1}{3}$$

B
$$\dfrac {1}{\sqrt {3}}$$

C
$$\dfrac {1}{\sqrt {2}}$$

D
$$\dfrac {2\sqrt {2}}{3}$$

E
$$\dfrac {2}{3\sqrt {2}}$$

Solution

Given

major axis of an ellipse $$=3\times$$ minor axis

$$a=3b$$

Eccentricity $$(e)=\sqrt{\dfrac{a^2-b^2}{a^2}}$$

$$e=\sqrt{\dfrac{(3b)^2-b^2}{(3b)^2}}=\sqrt{\dfrac{8b^2}{9b^2}}$$

$$e=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt 2}{3}$$

Two eccentricity of ellipse is $$\dfrac{2\sqrt 2}{3}$$
Question 10
1 / -0

If the latus-rectum of an ellipse is one half of its minor axis, then its eccentricity is

A
$$\dfrac {1}{2}$$

B
$$\dfrac {1}{\sqrt {2}}$$

C
$$\dfrac {\sqrt {3}}{2}$$

D
$$\dfrac {\sqrt {3}}{4}$$

Solution

Consider the equation of the ellipse is $$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1$$

It is given that; length of latus rectum = half of minor axis

$$\dfrac {2b^2}{a}=b \Rightarrow a=2b$$

now, $$b^2=a^2 (1-e^2)$$

$$\Rightarrow \ b^2=4b^2 (1-e^2)\Rightarrow 1-e^2 =1/4 \Rightarrow e^2 =3/4$$

$$\therefore \ $$ eccentricity $$(e)=\sqrt 3 /2$$

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Conic Sections Test - 48

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Conic Sections Test - 48

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Question 1

$$x ^ { 2 } + y ^ { 2 } + 4 x - 2 y + 4 = 0$$

$$x ^ { 2 } + y ^ { 2 } - 4 x + 2 y - 4 = 0$$

$$x ^ { 2 } + y ^ { 2 } + 4 x + 2 y - 4 = 0$$

$$x ^ { 2 } + y ^ { 2 } + 2 x - 4 y - 4 = 0$$

Solution

Question 2

$$3:4$$

$$3:2$$

$$5:4$$

$$5:3$$

Solution

Question 3

$$9$$

$$\dfrac {7}{\sqrt {5}}$$

$$\dfrac {9}{\sqrt {5}}$$

$$\dfrac {8}{\sqrt {5}}$$

Solution

Question 4

$$\lambda < 5$$

$$\lambda < 2$$

$$2 < \lambda < 5$$

$$\lambda < 2$$ or $$\lambda < 5$$

Solution

Question 5

$$25/16$$

$$4/5$$

$$16/25$$

$$5/4$$

Solution

Question 6

$$\dfrac 23$$

$$\dfrac 34$$

$$\dfrac 45$$

$$\dfrac 12$$

Solution

Question 7

$$\dfrac {(x + 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$$

$$\dfrac {(x - 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$$

$$\dfrac {(x - 1)^{2}}{16} + \dfrac {(y + 1)^{2}}{64} = 1$$

$$\dfrac {(x + 1)^{2}}{64} + \dfrac {(y - 1)^{2}}{16} = 1$$

Solution

Question 8

$$\dfrac 35$$

$$\dfrac 13$$

$$\dfrac 25$$

$$\dfrac 15$$

Solution

Question 9

$$\dfrac {1}{3}$$

$$\dfrac {1}{\sqrt {3}}$$

$$\dfrac {1}{\sqrt {2}}$$

$$\dfrac {2\sqrt {2}}{3}$$

$$\dfrac {2}{3\sqrt {2}}$$

Solution

Question 10

$$\dfrac {1}{2}$$

$$\dfrac {1}{\sqrt {2}}$$

$$\dfrac {\sqrt {3}}{2}$$

$$\dfrac {\sqrt {3}}{4}$$

Solution

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$$

x ^ { 2 } + y ^ { 2 } + 4 x - 2 y + 4 = 0

$$

$$

x ^ { 2 } + y ^ { 2 } - 4 x + 2 y - 4 = 0

$$

$$

x ^ { 2 } + y ^ { 2 } + 4 x + 2 y - 4 = 0

$$

$$

x ^ { 2 } + y ^ { 2 } + 2 x - 4 y - 4 = 0

$$