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Question

If a normal to the hyperbola x2a2y2b2=1 meets the axes at M & N and the lines MP & NP are drawn perpendicular to the axes meeting at P, then locus of P is

A
(a2x2b2y2)=(a2b2)
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B
a2x2b2y2=a2+b2
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C
a2x2b2y2=(a2+b2)2
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D
a2x2b2y2=(a2b2)2
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Solution

The correct option is C a2x2b2y2=(a2+b2)2
The equation of any normal to the hyperbola x2a2y2b2=1 is axcosϕ+by cotϕ=a2+b2(1)

Let the normal meets x -axis at M & y - axis at N respectivley, So
M=((a2+b2a)secϕ,0)&N=(0,(a2+b2b)tanϕ)

Let locus of P be (α,β)
Since PM&PN are perpendicular to the axes, co - ordinates of P is
((a2+b2a)secϕ,(a2+b2b)tanϕ)
α=(a2+b2a)secϕ & β=(a2+b2b)tanϕ
α(aa2+b2)=secϕ & β(ba2+b2)=tanϕ

As sec2ϕtan2ϕ=1
α2(aa2+b2)2β2(ba2+b2)2=1
α2a2β2b2=(a2+b2)2
Hence locus of (α,β) is
a2x2b2y2=(a2+b2)2

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