Mathematics - Ellipse Question with Solution | TestHub

Point of intersection of tangents at P and Q is $\text{R}\left(\text{2 sec}\theta ,0\right)$
Area of $\Delta \text{PQR}=\frac{1}{2}·2\sqrt{3}sin\theta ·\left(\text{2 sec}\theta -\text{2 cos }\theta \right)$
$\Rightarrow \Delta =2\sqrt{3}·\frac{{\text{sin}}^3\theta }{\text{cos }\theta }$; where $\text{cos }\theta \in \left[\frac{1}{4}\text{,}\frac{1}{2}\right]$

Now $\frac{\text{d}\Delta }{\text{d}\theta }=\frac{2\sqrt{3}\left|\text{cos }\theta ·{\text{3 sin}}^2\theta \text{cos }\theta -{\text{sin}}^3\theta \left(-\text{sin}\theta \right)\right|}{{\text{cos }}^2\theta }>0$
As $\theta$ increases, $\Delta$ increases $\Rightarrow$ when cos $\theta$ decreases, $\Delta$ increases
$\therefore {\Delta }_{\text{min}}\text{ occurs at cos}\theta =1/2\text{,}\text{Therefore }{\Delta }_2=2\sqrt{3}·\frac{{\left(1-1/4\right)}^{3/2}}{1/2}=4\sqrt{3}·\frac{3\sqrt{3}}{8}=\frac{36}{8}$
${\Delta }_{\text{max}}\text{ occurs at cos}\theta =1/4\text{,}\text{Therefore }{\Delta }_1=2\sqrt{3}·\frac{{\left(1-1/16\right)}^{3/2}}{1/4}=8\sqrt{3}·\frac{15\text{.}\sqrt{15}}{\text{4.4.4}}=\frac{2\sqrt{3}\text{.}15\text{.}\sqrt{3}\sqrt{5}}{16}$
$\Rightarrow {\Delta }_1=\frac{45}{8}\sqrt{5}$
Now $\frac{8}{\sqrt{5}}{\Delta }_1-8{\Delta }_2=45-36=9$