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Equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0) is

A. \frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9}

B. \frac{x^{2}}{9}-\frac{y^{2}}{9}=\frac{4}{9}

C. \frac{x^{2}}{4}-\frac{y^{2}}{9}=1

D. None of these
 

Option (A) is correct.

Let the equation of the hyperbola be \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1  

 e=3/2 

 Foci=(±ae),0=(±2,0)

  So, after comparing the equations, ae=2  

a*3/2=2  

 a=4/3 

  b2=a2(e2-1)   

 b2=(4/3)2((3/2)2-1) 

=(16/9)(9/4-1)

=16/9*5/4=20/9   

  Equation of hyperbola is \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

=\frac{x^{2}}{\left (\frac{4}{3} \right )^{2}}-\frac{y^{2}}{\frac{20}{9}}=1

\frac{x^{2}}{\frac{16}{9}}-\frac{y^{2}}{\frac{20}{9}}=1

  Hence, \frac{x^{2}}{16}-\frac{y^{2}}{{20}}=\frac{1}{9} is the required equation. 

 

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The distance between the foci of a hyperbola is 16 and its eccentricity is \sqrt{2}. Its equation is
A. x2 – y2 = 32
B. \frac{x^{2}}{4}-\frac{y^{2}}{9}=1
C. 2x – 3y2 = 7
D. none of these

Option (A) is correct.

Let the equation of the hyperbola be \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

Given that Foci=(±ae,0)

Distance between foci=2ae=16

 2*a*\sqrt{2}=16

a=4\sqrt{2}

b^{2}=a^{2}(e^{2}-1)=(4\sqrt{2})^{2}((\sqrt{2})^{2}-1)

=32(2-1)=32  

 Hence, equation is \frac{x^{2}}{32}-\frac{y^{2}}{32}=1

 

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The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is
A. 4/3

B. \frac{4}{\sqrt{3}}
C. \frac{2}{\sqrt{3}}
D. none of these

Let the equation of the hyperbola be x2/a2-y2/b2=1

length of latus rectum=8

2b2/a=8

b2=4a 

 Conjugate axis=half of the distance between the foci

2b=ae

b2=a2(e2-1)

\frac{a^{2}e^{2}}{4}=a2(e2-1)

e2=4/3

e=\frac{2}{\sqrt{3}}

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If e is the eccentricity of the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\left ( a<b \right ), then
A. b2 = a2 (1 – e2)
B. a2 = b2 (1 – e2)
C. a2 = b2 (e2 – 1)
D. b2 = a2 (e2 – 1)

Option (b) is correct.

Given that \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

We know that a2=b2(1-e2)

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 The length of the latus rectum of the ellipse 3x2 + y2 = 12 is
A. 4
B. 3
C. 8
D. 12

 

Option (d) is correct.

Given ellipse is 3x2+y2=12

\frac{x^{2}}{4}+\frac{y^{2}}{12}=1

a2=4

a=2

b2=12

b=2\sqrt{3}

length of latus rectum=\frac{2a^{2}}{b}=\frac{4}{\sqrt{3}}

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The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 = 0 and eccentricity 1/2 is
A. 7x2 + 2xy + 7y2 – 10x + 10y + 7 = 0
B. 7x2 + 2xy + 7y2 + 7 = 0
C. 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0
D. none

Option (a) is correct.

Given that focus of the ellipse is S(1,-1)and the equation of directrix is x-y-3=0  

Also, e=1/2 

From definition of ellipse, for any point P (x,y) on the ellipse, we have SP=ePM, where

 M is foot of the perpendicular from point P to the directrix. 

\sqrt{(x-1)^{2}+(y+1)^{2}}=\frac{1}{2}\frac{\left | x-y-3 \right |}{\sqrt{2}}

7x2+7y2+2xy-10x+10y+7=0

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If the vertex of the parabola is the point (–3, 0) and the directrix is the line x + 5 = 0, then its equation is
A. y2 = 8 (x + 3)
B. x2 = 8 (y + 3)
C. y2 = – 8 (x + 3)
D. y2 = 8 ( x + 5)

Given: Vertex =(–3, 0) and the directrix x = -5(M)

So focus S (-1,0)  

For any point of parabola P (x,y) we have 

SP=PM

\sqrt{(x+1)^{2}+y^{2}}=\left | x+5 \right |

x2+y2+2x+1=x2+10x+25

y2=8x+24

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If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus rectum is
A. 2/3

B. 4/3
C. 1/3
D. 4

Option (b) is correct.

Parabola y2=4ax, passes through the point (3,2)

4=4a(3)

a=1/3

Length of latus rectum 4a=4/3

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If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is
A. x2 = –12y
B. x2 = 12y
C. y2 = –12x
D. y2 = 12x

a Given that, focus of parabola is at S(0,-3)and equation of directrix is y=3 

 For any point P(x,y)on the parabola, we have

 SP=PM

\sqrt{(x-0)^{2}+(y+3)^{2}}=\left | y-3 \right |

x2+y2+6y+9=y2-6y+9

x2=-12y

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The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
A. x2 + y2 = 9a2
B. x2 + y2 = 16a2
C.x2 + y2 = 4a2
D. x2 + y2 = a2

Option (C)

median length=3a

Radius of circle=2/3*median length=2a

Equation of circle

x2+y2=4a2

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