Question

A vertical line passing through the point $(h,0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta \left(h\right)=\text{area of the triangle}PQR$, ${\Delta }_1=\underset{1/2\le h\le 1}{max}\Delta \left(h\right)$ and ${\Delta }_2=\underset{1/2\le h\le 1}{min}\Delta \left(h\right)$, then $\frac{8}{\sqrt{5}}{\Delta }_1-8{\Delta }_2=$

Answer 1

Equation of the line segment $PQ$ is $x=h$
$PQ$ is the chord of contact.
$\Rightarrow \frac{x\alpha }{4}+0=1$
Comparing the two equations, we get
$\alpha =\frac{4}{h}$

$\frac{h^2}{4}+\frac{y^2}{3}=1\Rightarrow y=\sqrt{3}\sqrt{1-\frac{h^2}{4}}$

$\therefore$ Coordinates of $P$ and $Q$ are $(h,\sqrt{3}\sqrt{1-\frac{h^2}{4}})$ and $(h,-\sqrt{3}\sqrt{1-\frac{h^2}{4}})$ respectively.

So, $\Delta \left(h\right)=\frac{1}{2}\times 2\sqrt{3}\sqrt{1-\frac{h^2}{4}}\times (\frac{4}{h}-h)$

i.e., $\Delta \left(h\right)=\frac{\sqrt{3}}{2}\times \frac{(4-h^2{)}^{3/2}}{h}$

$\frac{d\Delta }{dh}=-\frac{\sqrt{3}(h^2+2)\sqrt{4-h^2}}{h^2}$

$\frac{d\Delta }{dh}<0\forall h\in [\frac{1}{2},1]$
$\Rightarrow \Delta$ is strictly decreasing.

So, ${\Delta }_1=\Delta \left(\frac{1}{2}\right)=\frac{\sqrt{3}}{8}(15{)}^{3/2}=\frac{\sqrt{3}}{8}\times 15\times \sqrt{15}=\frac{45\sqrt{5}}{8}$

and ${\Delta }_2=\Delta \left(1\right)=\frac{9}{2}$

$\therefore \frac{8}{\sqrt{5}}{\Delta }_1-8{\Delta }_2=45-36=9$