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 the tangents at $P$ and $Q$ of the ellipse $x^2+2y^2=6$ are at right angles

Answer 1

Given, $x^2+4y^2=4$ or $\frac{x^2}{4}+\frac{y^2}{1}=1\dots$ (i)
Equation of any tangent to the ellipse on (i) can be written as
$\frac{x}{2}\cos \theta +y\sin \theta =\mathrm{1...}$ (ii)
Equation of second ellipse is

$x^2+2y^2=6$
$\Rightarrow \frac{x^2}{6}+\frac{y^2}{3}=\mathrm{1....}$(iii)
Suppose the tangents at $P$ and $Q$ meets at $A(h,k)$. Equation of the chord of contact of the tangents through $A(h,k)$ is
$\frac{hx}{6}+\frac{ky}{3}=\mathrm{1....}$ (iv)
But Eqs. (iv) and (ii) represent the same straight line, so comparing Eqs. (iv) and (ii), we get
$\frac{h/6}{\cos \theta /2}=\frac{k/3}{\sin \theta }=\frac{1}{1}$
$\Rightarrow h=3\cos \theta$ and $k=3\sin \theta$
Therefore, coordinates of $A$ are $(3\cos \theta ,3\sin \theta )$.
Now, the joint equation of the tangents at $A$ is given by $T^2=S{S}_1$,
i.e. ${\left(\frac{hx}{6}+\frac{ky}{3}-1\right)}^2=\left(\frac{x^2}{6}+\frac{y^2}{3}-1\right)\left(\frac{h^2}{6}+\frac{k^2}{3}-1\right)\dots$ (v)
In Eq. (v), coefficient of $x^2=\frac{h^2}{36}-\frac{1}{6}\left(\frac{h^2}{6}+\frac{k^2}{3}-1\right)$
$=\frac{h^2}{36}-\frac{h^2}{36}-\frac{k^2}{18}+\frac{1}{6}=\frac{1}{6}-\frac{k^2}{18}$
and coefficient of $y^2=\frac{k^2}{9}-\frac{1}{3}\left(\frac{h^2}{6}+\frac{k^2}{3}-1\right)$
$=\frac{k^2}{9}-\frac{h^2}{18}-\frac{k^2}{9}+\frac{1}{3}=-\frac{h^2}{18}+\frac{1}{3}$
Again, coefficient of $x^2+$ coefficient of $y^2$
$=-\frac{1}{18}\left(h^2+k^2\right)+\frac{1}{6}+\frac{1}{3}$
$=-\frac{1}{18}\left(9{\cos }^2\theta +9{\sin }^2\theta \right)+\frac{1}{2}=-\frac{9}{18}+\frac{1}{2}=0$
which shows that two lines represent by Eq. (v) are at right angles to each other.