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Question

If the eccentricity of a hyperbola is 2 and if the distance between the foci is 16, then its equation is


A

x2y2=4

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B

x2y2=8

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C

x2y2=16

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D

x2y2=32

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Solution

The correct option is D

x2y2=32


Given that distance between foci, 2c =16
c=8; e=2
a=ce=82=42b=c2a2=32=42

The equation of the hyperbola is
x232y232=1x2y2=32


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