The locus of the mid point of normal chord of ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ (A) $x^2+y^2=2$
(B) $x^2-y^2=1$
(C) $(\frac{a^6}{x^2}+\frac{b^6}{y^2})(\frac{x^2}{a^2}+\frac{y^2}{b^2})^2=(a^2-b^2)^2$
(D) $x^2+y^2=1$
How I tried
I can't type the rough work here because its huge, but the idea was to assume mid point of normal as $(h,k)$, then use $T=S_1$ to find equation of chord then solve the chord and ellipse to get point of contact, then find slope of tangent and slope of normal from that point and use $m_1m_2=-1$, but after few steps the calculation was among bi-quadratic, and very complex things that was very difficult to substitute.
The answer given is (C).
So my question is how would you solve the problem ? Is there a clean and easy way than my method mentioned above, so that I can use it in examination hall to get answer quickly.
Since it not a circle nor hyperbola it is c)