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Question

If the equation to the hyperbola is 5x2−12xy+5y2+22x−26y+29=0 and the equation of pair of asymptotes is 5x2−12xy+5y2+22x−26y+26611=0 then the equation of conjugate hyperbola is

A
5x212xy+5y2+22x26y+58=0
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B
5x212xy+5y2+22x26y+10611=0
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C
5x212xy+5y2+22x26y21311=0
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D
5x212xy+5y2+22x26y+21311=0
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Solution

The correct option is D 5x212xy+5y2+22x26y+21311=0
Given: Hyperbola 5x212xy+5y2+22x26y+29=0; Pair of asymptotes 5x212xy+5y2+22x26y+26611=0

To Find: Equation of conjugate hyperbola

We know that, on adding the equation of a hyperbola and its conjugate hyperbola results into twice the equation of pair of asymptotes.

5x212xy+5y2+22x26y+29+(Conjugate hyperbola)=2(5x212xy+5y2+22x26y+26611)

Conjugate hyperbola=5x212xy+5y2+22x26y+21311

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