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Question

The equation of the conjugate hyperbola of the hyperbola x22y225x42y3=0 is

A
x22y225x42y+2=0
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B
x22y225x42y+5=0
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C
x22y225x42y+3=0
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D
x22y225x42y2=0
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Solution

The correct option is B x22y225x42y+5=0
Conjugate hyperbola for x2a2y2b2=1 is x2a2y2b2=1

Equation can be written as (x225x+5)2(y2+22y+2)=3+5+(4)
(x5)24(y+2)22=1

Conjugate hyperbola is (x5)24(y+2)22=1
x22y225x42y+5=0

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