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 $-1$.
Best Answer
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.