Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse,$\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{2}=1$ from any of its foci?

We know that the locus of the feet of the perpendicular drawn from foci to any tangent of the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ is the auxiliary circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$.
$\therefore$ Auxiliary circle  is $x^2+{\mathrm{y}}^2=4$.
$\therefore (-1,\sqrt{3})$ ) satisfies the given equation.