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Question

A tangent to x2a2+y2b2=1 meets the axes at A and B. Then the locus of mid point of AB is

A
a2x2+b2y2=2
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B
a2x2+b2y2=4
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C
a2x2+b2y2=1
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D
a2x2b2y2=12
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Solution

The correct option is B a2x2+b2y2=4
Let equation of tangent to the given ellipse is xx1a2+yy1b2=1
Finding its intersection with X and Y axes
A=(a2x1,0);B=(0,b2y1)
Mid point is (a22x1,b22y1)=(α,β)
(x1,y1) lies on Ellipse
a24α2+b24β2=1
a2x2+b2y2=4 is equation of required locus.

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