<s>[INST] 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?
 [/INST] $$x(a\sin t)-y(b\cos t)=0\ \ \ \  (1)\implies\tan t=\dfrac{by}{ax}\ \ \ \ (3)$$
$$x(b\sin t)+ya(e-\cos t)-abe\sin t=0\ \ \ \  (2)$$
$$\dfrac x{ab^2e\sin t\cos t}=\dfrac y{a^2be\sin^2t}=\dfrac1{a^2e\sin t(1-e\cos t)}$$
$$\implies\dfrac{1-e\cos t}{\cos t}=\dfrac{b^2}{ax}\implies\sec t=\dfrac{b^2}{ax}+e\ \ \ \ (4)$$
Use $\sec^2t-\tan^2t=1$ to eliminate $t$
$$\left(\dfrac{b^2}{ax}+e\right)^2-\left(\dfrac{by}{ax}\right)^2=1$$
$$\iff(ax)^2+(by)^2=(b^2+aex)^2$$
$$\iff b^4+2ab^2ex=b^2y^2+a^2x^2(1-e^2)$$
Use $b^2=a^2(1-e^2)$
$$b^2(b^2+2aex)=b^2(y^2+x^2)$$
</s>