Locus of a point of intersecting lines

Question

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?

Answer 1

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$

Answer 2

$$x(a\sin t)-y(b\cos t)=0\ \ \ \ (1)\implies\tan t=\dfrac{by}{ax}\ \ \ \ (3)$$

$$x(b\sin t)+ya(e-\cos t)-abe\sin t=0\ \ \ \ (2)$$

$$\dfrac x{ab^2e\sin t\cos t}=\dfrac y{a^2be\sin^2t}=\dfrac1{a^2e\sin t(1-e\cos t)}$$

$$\implies\dfrac{1-e\cos t}{\cos t}=\dfrac{b^2}{ax}\implies\sec t=\dfrac{b^2}{ax}+e\ \ \ \ (4)$$

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 $b^2=a^2(1-e^2)$

$$b^2(b^2+2aex)=b^2(y^2+x^2)$$