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Question

Tangent at any point on the hyperbola x2a2y2b2=1 cut the axes at A and B respectively. If the rectangle OAPB (where O is origin) is completed then locus of point P is given by

A
a2x2b2y2=1
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B
a2x2+b2y2=1
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C
a2y2b2x2=1
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D
none of these
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Solution

The correct option is C a2x2b2y2=1

Consider the equation of the parabola.

x2a2y2b2=1

We know that the equation of the tangent from any point M(asecθ,bsinθ) to the hyperbola is,

xsecθaytanθb=1

Therefore,

Coordinates of A=(0,btanθ)

Coordinates of B=(asecθ,0)

Now, consider rectangle OAPB. Let the point be P(h,k). Then,

AP=OB

h=asecθ

secθ=ah

PB=OA

k=btanθ

tanθ=bk

We know that,

sec2θtan2θ=1

a2h2b2k2=1

Now, substitute (x,y) for (h,k).

a2x2b2y2=1

Hence, this is the required locus of the point.

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