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

Parameterise $x^2+4y^2=4$ with $x=2\cos t$, $y=\sin t$
Tangent at $t$ is $\Rightarrow x\left(2\cos t\right)+4y\left(\sin t\right)=4\Rightarrow x\cos t+2y\sin t=2$ … $\left(i\right)$
The coordinates of $P,Q$ are the solutions of $\left(i\right)$ and $x^2+2y^2=6$ … $\left(ii\right)$
Differentiating w.r.t $x$

$\Rightarrow 2x+4y{y}^{\prime }=0$

So gradient at $(x,y)$ is $m=–\frac{x}{2y}$ … $\left(iii\right)$

Tangent gradients at $P,Q$ $\left(m\right)$ are found by eliminating $x,y$ from $\left(i\right),\left(ii\right)and\left(iii\right)$
Using $x$ from $\left(iii\right)$ in $\left(i\right)and\left(ii\right)$ gives $\Rightarrow \frac{1}{y}=\sin t-m\cos t$ and $\frac{3}{y^2}=2m^2+1$
$\Rightarrow 3(\sin t-m\cos t{)}^2=2m^2+1$

$\Rightarrow m^2(3{\cos }^{2}t-2)–6m\cos t\sin t+(3{\sin }^{2}t-1)=0$
Because $\Rightarrow {\cos }^{2}t+{\sin }^{2}t=1$ this is $m^2(3{\cos }^{2}t-2)–6m\cos t\sin t+(2-3{\cos }^{2}t)=0$
By properties of quadratic, it follows that $m_1{m}_2=-1$ and so tangents are perpendicular.