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Question

Let P,D be two points on the ellipse x2a2+y2b2=1, whose eccentric angles differ by π2. Then the locus of mid point of chord PD is

A
x2a2+y2b2=2
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B
x2a2+y2b2=4
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C
x2a2+y2b2=14
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D
x2a2+y2b2=12
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Solution

The correct option is D x2a2+y2b2=12
Given ellipse is x2a2+y2b2=1
Let P(acosθ,bsinθ), then
D(acos(θ+π2),bsin(θ+π2))
D(asinθ,bcosθ)
Let mid point of PD be M(h,k)
So, h=acosθasinθ2
cosθsinθ=2ha (1)
And
k=bsinθ+bcosθ2
sinθ+cosθ=2kb (2)
Squaring and adding equation (1) and (2)
2sin2θ+2cos2θ+2sinθcosθ2sinθcosθ=(2ha)2+(2kb)2
2=(2ha)2+(2kb)2
x2a2+y2b2=12

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