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.

Question

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.

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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.

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

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

Answer 1

It is well known that the locus of points where two perpendicular tangent to the ellipse cross each other is a circle, called the director circle of the ellipse, whose radius is $r=\sqrt{a^2+b^2}=3$ in our case.

We can then revert the proof: taken any point $(h,k)$ on the director circle (hence $h^2+k^2=9$), its chord of contact $PQ$ to ellipse $E_2$ has the simple equation $$ {hx\over6}+{ky\over3}=1 $$ and it then suffices to show that this line is tangent to ellipse $E_1$ for all values of $(h,k)$. In fact, substituting $y$ from the above equation into the equation of $E_1$ yields the resolvent equation $$ \left({3\over2}x-h\right)^2=0 $$ with a single solution $x=2/3 h$. That completes the proof.

Answer 2

I'd like to solve the converse problem (given tangents to $P$ and $Q$ intersects at right angle then line $PQ$ has exactly $1$ intersection point with inner ellipse) and revert the solution. Let $P(x_p,y_p)$ and $Q(x_q,y_q)$.
Performing implicit differentiation on $x^2+2y^2=6$ we have $$2x+4yy'=0\Rightarrow y'=\frac{-x}{2y},\,y\ne 0.$$ Then the tangets to the larger ellipse at $P,Q$ are, resp: $$\frac{-x_p}{2y_p}(x-x_p)=(y-y_p)\hbox{ and } \begin{cases} x=x_q+2y_qt\\ y=y_q-x_qt \end{cases}$$ The intersection: $$-x_p(x_q+2y_qt-x_p)=2y_p(y_q-x_qt-y_p)$$ $$y_p x_q - x_p y_q\ne0,\, t = \frac{-x_p^2 + x_p x_q - 2 y_p^2 + 2 y_p y_q}{2 (y_p x_q - x_p y_q)}$$ and the intersection point is $I=(x_q+2y_qt,y_q-x_qt)$. Vectors $PI$ and $QI$, resp: $$PI=(x_q-x_p+2y_qt,y_q-y_p-x_qt),\,QI=t(2y_q,-x_q)$$ And the inner (dot) product is $0$: $$(x_q-x_p+2y_qt)\cdot 2y_q-(y_q-y_p-x_qt)x_q=0\Leftrightarrow$$ $$\frac{(x_p x_q + 4 y_p y_q) (x_p x_q + 2 y_p y_q - x_q^2 - 2 y_q^2)}{2 (x_p y_q - y_p x_q)}=0$$ Now we take $R$ on $PQ$: $R=uP+(1-u)Q$ and the intersection $x_R^2+4y_R^2=4$: $$(ux_p+(1-u)x_q)^2+4(uy_p+(1-u)y_q)^2-4=0$$ is quadratic in terms of $u$ and it's discriminant $$-16 (x_p^2 y_q^2 - x_p^2 - 2 x_p y_p x_q y_q + 2 x_p x_q + y_p^2 x_q^2 - 4 y_p^2 + 8 y_p y_q - x_q^2 - 4 y_q^2)$$ indeed equals 0. Don't know how wolframalpha did get there.
Now we see the converse is true with the additional results of $u=\frac{-4}{x_p^2-12}$ and $x_I^2+y_I^2=9$.
Now the original problem seems more simple:
1. We take arbitrary $R$ on the inner ellipse
2. Make tangent in $R$ intersect the outer ellipse in $P,Q$ (quadratic)
then either 3. compute $u=\frac{|RQ|}{|PQ|}$ and compare with $\frac{-4}{x_P^2-12}$
or 4. Intersect the tangents to the outer ellipse at $P,Q$ with $x^2+y^2=9$ and see there are $3$ distinct points, with two of them symmetrical thruogh $O(0,0)$ so the angle at the third point is right as two others form a diameter of $x^2+y^2=9$.

Answer 3

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)):

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, $$\color{red}{x_1x_2 = \frac{4(1-3\sin^2 \theta)}{1+\sin^2\theta}=\frac{-4(2-3\cos^2 \theta)}{1+\sin^2\theta}}.$$

2. 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.