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Question

A hyperbola, having the transverse axis of length 2sinθ, is confocal with the ellipse 3x2+y2=12 then, its equation is


A

x2cosec2θy2sec2θ=1

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B

x2sec2θy2cosec2θ=1

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C

x2sin2θy2cos2θ=1

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D

x2cos2θy2sin2θ=1

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Solution

The correct option is A

x2cosec2θy2sec2θ=1


The explanation for the correct option:

Step 1: Expressing the given information:

The given ellipse is

3x2+y2=12x24+y23=1

Hence,

a2=4b2=3a=2b=3

Step 2: Calculating the variables for the equation

e=1-34=12

The focus of the ellipse = (±ae,0)=(±1,0)

Given required hyperbola is confocal to the ellipse

Let, a',b',e' be the transverse axis, conjugate axis, and eccentricity of the hyperbola

a'e'=1sinθ.e'=1e'=1sinθ

b'2=a'2(e21)b'2=1sin2θ=cos2θ

Hence the required hyperbola is

x2sin2θ-y2cos2θ=1

Hence, the equation of the hyperbola is x2cosec2θ-y2sec2θ=1 so, option (A) is the correct option.


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