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Question

A tangent 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
a2x2+b2y2=12
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Solution

The correct option is B a2x2+b2y2=4
Tangent to the ellipse is given by,

xcosmθa+ysinθb=1

Since this line intersect the cordinate axis
at y=0,xcosθa=1x=acosθ

so the cordinate is (acosθ,0)

Similarly
at x=0,ysinθb=1y=bsinθ

Cordinate is (0,bsinθ)

Midpoint is given by-

(h,k)=acosθ+02,0+bsinθ2

(h,k)=(a2cosθ,b2sinθ)

we get.
cosθ=a3h,sinθ=b2k

As we know
sin2θ+cos2θ=1a24h2+b24k2=1

a2h2+b2k2=4

therefore the locus of midpoint is

a2x2+b2y2=4




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