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Question

If normal at a variable point P on the ellipse x2a2+y2b2=1 of eccentricity e meets the axes of the ellipse at Q and R, then the locus of the midpoint of QR is a conic with an eccentricity e such that

A
e is independent of e
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B
e=1
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C
e=e
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D
e=e2
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Solution

The correct option is C e=e
Assuming a>b for the given ellipse
Normal at point P(acosθ,bsinθ) is given by :
axcosθbysinθ=a2b2(i)
It meets coordinate axis at Q((a2b2)cosθa,0) and R(0, (a2b2)sinθb)
Let T(h,k) be the midpoint of QR, then
2h=(a2b2)cosθa and
2k=(a2b2)sinθb
cos2θ+sin2θ=4h2a2(a2b2)2+4k2b2(a2b2)2=1
Hence, required locus is
x2(a2b2)24a2+y2(a2b2)24b2=1
which is an ellipse, having eccentricity e, then
e=       1((a2b2)24a2)((a2b2)24b2)=1b2a2
e=e

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