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