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Question

A normal to the hyperbola x2a2y2b2=1 meets the axes in M and N and lines MP and NP are drawn perpendiculars to the axes meeting at P. Then the locus of P is :

A
a2x2b2y2=(a2b2)2
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B
a2x2b2y2=(a2+b2)2
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C
b2y2a2x2=(a2b2)2
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D
b2y2a2x2=(a2+b2)2
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Solution

The correct option is C a2x2b2y2=(a2+b2)2
Equation of a normal at a parametric point is atanθ(xasecθ)+bsecθ(ybtanθ)=0
Thus, M=(0,a2tanθ+b2tanθb)
N=(a2secθ+b2secθa,0)
Thus, point P is given by (a2secθ+b2secθa,a2tanθ+b2tanθb)
Now, let x=a2secθ+b2secθa and y=a2tanθ+b2tanθb
secθ=axa2+b2 and tanθ=bya2+b2
Using the identity sec2θtan2θ=1, we get a2x2b2y2=(a2+b2)2
Hence, option 'B' is correct.

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