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Question

# Tangent at any point on the hyperbola x2a2âˆ’y2b2=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 a2x2−b2y2=1Consider the equation of the parabola. x2a2−y2b2=1 We know that the equation of the tangent from any point M(asecθ,bsinθ) to the hyperbola is, xsecθa−ytanθb=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, sec2θ−tan2θ=1 a2h2−b2k2=1 Now, substitute (x,y) for (h,k). a2x2−b2y2=1 Hence, this is the required locus of the point.

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