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Question

If 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
none of these
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Solution

The correct option is B a2x2+b2y2=4
Let the equation of tangent to the ellipse x2a2+y2b2=1 be xacosθ+ybsinθ=1
So, the points
A=(acosθ,0), B=(0,bsinθ)
Let the mid point of AB be (h,k)
a2cosθ=h, b2sinθ=k
2cosθ=ah,2sinθ=bk
The required locus is a2x2+b2y2=4

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