Locus of a point of intersecting lines The perpendicular from the centre of the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ to the tangent at any point $P$ meets the line joining  point $P$ to the focus $(ae,0)$ at $Q$. Prove that the locus of $Q$ is $$x^2+y^2-2aex=b^2$$
My attempt
Tangent to an ellipse at $P(\theta)$
$$bx\cos\theta +by\sin\theta=ab\tag{1}$$
slope $m_1$ of a normal line to $(1)$
$$m_1=\frac{a\sin\theta}{b\cos\theta}$$
Equation of the normal passing through origin
\begin{align*}
y=m_1x
&=\frac{a\sin\theta}{b\cos\theta}x\tag{2}
\end{align*}
Equation of a line through $P$ and $S$
$$\begin{align*}
y=m_2(x-ae)&=\frac{b\sin\theta}{\left(a\cos\theta-ae\right)}(x-ae)
\tag{3}
\end{align*}$$
Point $Q(h,k)$ satisfies $(2)$ and $(3)$.
How to eliminate $\sin\theta$ and $\cos\theta$ to obtain the required locus?

You are almost there. Now equating both,
$\displaystyle \frac{a}{b\cos\theta} x = \frac{b}{\left(a\cos\theta-ae\right)} (x-ae)$
$(x-ae+ae) (a^2\cos\theta-a^2e) = b^2\cos\theta(x-ae)$
$(x-ae) (a^2 \cos \theta - b^2 \cos\theta - a^2e)= a^3e^2 - a^3e \cos\theta$
We will use the fact that $a^2 - b^2 = a^2e^2$ now and again later.
$x-ae = \displaystyle \frac{a\cos\theta-ae}{1 - e \cos\theta}$
$x = \displaystyle \frac{b^2\cos\theta}{a(1-e\cos\theta)}$
$y = \displaystyle \frac{b \sin\theta}{1-e\cos\theta}$
$x^2 + y^2 - 2aex = (x-ae)^2 + y^2 - a^2e^2$
$ = \displaystyle \frac{a^2 \cos^2\theta+a^2e^2-2a^2e\cos\theta + (a^2-a^2e^2)\sin^2\theta}{(1-e\cos\theta)^2} - a^2e^2$
$ = a^2 - a^2e^2 = b^2$
