In the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the parametric equation for a tangent is $\frac{x}{a}\cos(h)+\frac{y}{b}\sin(h)=1$, and the foci are $(ae,0)$ and $(-ae,0)$, taking the focus on the positive $x$-axis, we get the following equation for the perpendicular:
$$\frac{x}{b}\sin(h)-\frac{y}{a}\cos(h)-\frac{ae}{b}\sin(h)=0$$
Solving these, we get the following coordinates for the foot of the perpendicular:
$$\left(a\cos(h),\frac{b\sin(h)}{1+e\cos(h)}\right)$$ which doesn't lie on the auxiliary circle for any value of $h$ other than $nπ$, where $n∈W$, which if because those are the only points where the foot of the perpendicular and the point of contact have the same ordinate, their abcissa is always the same ($a\cos(h)$).
Best Answer
The parametric equation of the line perpendicular to the tangent through the focus is $$\frac{(x \sin(h))}{b} - \frac{(y \cos(h))}{a} = \frac{ae \sin(h)}{b}.$$
Now, we need to find the intersection with the tangent line, hence squaring and adding the two equations of the tangent and perpendicular lines (to try to eliminate the parameter $h$) we get $$x^2 \biggl[\frac{\cos(h)^2}{a^2} + \frac{\sin(h)^2}{b^2}\biggr] + y^2 \biggl[\frac{\cos(h)^2}{a^2} + \frac{\sin(h)^2}{b^2}\biggr] = \frac{a^2 \sin(h)^2 + b^2 \cos(h)^2}{b^2}$$ [Write $e^2$ as $(a^2-b^2)/a^2$].
On further simplification, we get the locus as the auxiliary circle: $$x^2 + y^2 = a^2$$