A tangent to ellipse $E_1$ intersects ellipse $E_2$ at $P$ and $Q$. Prove that the tangents to $E_2$ at $P$ and $Q$ are perpendicular to each other. A tangent to the ellipse $x^2 + 4y^2 = 4$ meets the ellipse $x^2 + 2y^2 = 6$ at P and Q. Prove that tangents at $P$ and $Q$ of ellipse $x^2 + 2y^2 = 6$ are perpendicular to each other. 

Let $E_1: \frac{x^2}{4} + \frac{y^2}{1} = 1$ and $E_2: \frac{x^2}{6} + \frac{y^2}{3} = 1$ be the equation of the ellipses. 
Let $R(h, k)$ be the point of tangent on $E_1$
Then let $L_1: y - k = m(x - h)$ be a line passing through two points $P$ and $Q$ on the outer ellipse. This line is tangential to $E_1$ at $R$.
To find the slope of the tangent at $R$ we take the derivative of $E_1$ at $R$. 
$$m = \frac{-h}{4k}$$
$$y - k = \frac{-h}{4k}(x - h)$$
$$ h^2 + 4k^2 = 4ky + hx \tag{1}$$
We know that $R(h, k)$ satisfies $E_1$. Hence, $$h^2 + 4k^2 = 4 \tag{2}$$
$(1)$ and $(2)$ implies, 
$$4ky + hx = 4 \tag{3}$$
$(3)$ is the equation of the tangent to $E_1$ at $R$.
I don't know how to proceed further.

I can think of various approaches, but can't work it out. 


*

*Proving that the point of intersection of the two tangents lies on a circle of diameter $PQ$. (We know that angle in a semicircle is $90$ degrees). 

*We can prove that product of slopes of the two tangents to $E_2$ is $-1$. 


Let $A=(2\cos \theta, \sin \theta)$ be the point of tangency on $E_1$. The slope of this tangent line will be 
$$t_A=-\frac{\cot \theta}{2}.$$
The equation of the tangent line $\ell$ at $A$ on ellipse $E_1$ is given by:
$$\ell: x\cos \theta+2y \sin \theta=2. \tag{1}$$
Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be the points on $E_2$ where the tangent $\ell$ intersects $E_2$.
Consider the intersection of $E_2$ ($x^2+2y^2=6$) and $\ell$ (equ (1)):


*

*We can get a quadratic equation in $x$ which will be of the form:
$$x^2(1+\sin^2\theta)-4x \sin \theta +4(1-3\sin^2\theta)=0.$$
This should have roots $x_1,x_2$ (the $x-$coordinates of $P$ and $Q$). Using Viete's relation,
$$\color{red}{x_1x_2 = \frac{4(1-3\sin^2 \theta)}{1+\sin^2\theta}=\frac{-4(2-3\cos^2 \theta)}{1+\sin^2\theta}}.$$


*We can get a quadratic equation in $y$ which will be of the form:
$$y^2(1+\sin^2\theta)-4y \cos \theta +(2-3\cos^2\theta)=0.$$
This should have roots $y_1,y_2$ (the $y-$coordinates of $P$ and $Q$). Using Viete's relation,
$$\color{blue}{y_1y_2 = \frac{(2-3\cos^2 \theta)}{1+\sin^2\theta}}=\color{green}{\frac{-x_1x_2}{4}}.$$
Slope of tangents at $P$ and $Q$ are
$$t_P=\frac{-x_1}{2y_1} \quad \text{ and } \quad t_Q=\frac{-x_2}{2y_2}.$$
For these tangents to be perpendicular, we want to show that
$$t_p \cdot t_Q=-1 \implies \boxed{\frac{x_1x_2}{4y_1y_2}=-1}.$$ 
Plug in the expressions for $x_1x_2$ and $y_1y_2$ and you get the answer.
