JEE Main · 2014 · 06 Apr 2014 · Offline · Ellipse

Write the ellipse as $\frac{x^2}{6}+\frac{y^2}{2}=1$, so $a^2=6$ and $b^2=2$, and a tangent of slope $m$ is $y=mx\pm \sqrt{6m^2+2}$. Let $(h,k)$ be the foot of the perpendicular from the centre to this tangent; the radius to the foot is perpendicular to the tangent, so $m=-\frac{h}{k}$. Imposing that $(h,k)$ lies on the tangent and squaring, $6h^2+2k^2=(h^2+k^2{)}^2$. The locus is $(x^2+y^2{)}^2=6x^2+2y^2$.