From a point $(P)$ on auxiliary circle of an ellipse a perpendicular is dropped on major axis of ellipse. The perpendicular cuts ellipse at another point $(Q)$.
Now from one of the focus $(S)$ of ellipse a perpendicular is dropped on tangent at point $(R)$ on point $(P)$.
Prove that $\triangle QRS$ is isosceles.
Finding coordinates of points $Q,R,S$ is a lengthy process.
I need a different method to deal with this question.
Thanks in advance for any hints of solution.
Let us make a simple parametrization, so that the coordinates of the point $R$ are visible by simple means:
In the picture, $O$ is the intersection of the axes of the ellipse, then $OP\|SR$ being both perpendicular on the tangent in $P$ to the external (major) auxiliary circle of the ellipse. We choose $T$ such that $PRST$ is a rectangle. Then we can easily characterize the vector $\overset \to{PR}=\overset \to{TS}$ from the data in the triangle $\Delta TOS$ with hypotenuse $OS=c=\sqrt{a^ 2-b^2}$ by using the angle $\angle TOS=\angle POS$ as a parameter. Let $TU$ be its height. Then we list: $$ \begin{aligned} TO &= OS\cos t=c\cos t\ ,\\ TU &= TO\sin t=c\sin t\cos t\ ,\\ US &= TS\cos(90^\circ-t)=OS\sin t\cos(90^\circ-t)=c\sin^2 t\ ,\\ P &= (a\cos t,a\sin t)\ ,\\ Q &= (a\cos t,b\sin t)\ ,\\ R &= P+(US,-TU)=a(\cos t,\sin t)+c\sin t(\sin t,-\cos t)\ ,\\ &\qquad\text{(which can be written directly from the picture)}\\ S &= (c,0)\ ,\\ SR &=\|\ (a\cos t-c\cos^2 t, a\sin t-c\sin t\cos t)\ \|\\ &=(a-c\cos t)\;\|\ (\cos t, \sin t)\ \|\\ &=a-c\cos t\ ,\\ &\qquad\text{(or simpler $SR=TP=OP-OT=a-c\cos t$)}\\ SQ^2 &=\|\ (c-a\cos t,-b\sin t)\ \|^2\\ &= c^2-2ac\cos t + a^2\cos^2 t+b^2 \sin^2 t\\ &= c^2\cos^2 t-2ac\cos t + a^2\cos^2 t+ c^2\sin^2 t +b^2 \sin^2 t \\ &= c^2\cos^2 t-2ac\cos t + a^2\\ &= (a-c\cos t)^2\ . \end{aligned} $$
So $SQ=SR$.
$\square$
(The path to the solution can be simplified, some of the relations omitted, but i usually give "them" all. For instance, we do not need analytically the coordinates of $R$, it is enough to compute $SR=TP=OP-OT$...)