JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Ellipse

The foot of the perpendicular dropped from a focus of an ellipse onto any tangent always lies on the auxiliary circle, $x^2+y^2=a^2$. For $\frac{x^2}{4}+\frac{y^2}{2}=1$ we have $a^2=4$, so the locus is $x^2+y^2=4$. The point $(-1,\sqrt{3})$ gives $1+3=4$, so it lies on this circle and hence on the required locus.