Equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$

Find the equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$.

will it also be an ellipse?

I think it is a circle whose radius is $a$ and center is the origin.

As is shown in the graph, suppose C is a point on the ellipse, and F1,F2 are two focus, CM is the angle bisector of $\angle F_1CF_2$, and F2P is perpendicular to the tangent line, E lies at both line F1C and F2P.

According to the reflexive property, it is easy to know that CF2 = CE, and hence $\triangle CF_2P$ is congruent with $\triangle CEP$, so $P$ is the mid point of $F_2E$ and $O$ is the midpoint of $F_1F_2$, so $OP$ is always half the length of $F_1E$, and we know that $ CF_1 + CE = CF_1 + CF_2 = 2a$, so $OP = a$.

Foot of perpendicular drawn from focus to any tangent lies on the auxiliary circle of the ellipse hence the locus will be a circle and not an ellipse. Moreover you may also try to verify the property by finding the the foot of perpendicular and putting it into the equation: $$ x^2 + y^2 = a^2 \qquad\text{(equation of auxillary circle for a standard ellipse)} $$

where $a$ is the length of semi-major axis.