A vertical line passing through the point (h , 0) intersects the...

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

Now $\dfrac{\text{d} \Delta}{\text{d} \theta} = \dfrac{2 \sqrt{3} \left|\text{cos} \theta \cdot {\text{3 sin}}^{2} \theta \text{cos} \theta - {\text{sin}}^{3} \theta \left(- \text{sin} \theta\right)\right|}{{\text{cos}}^{2} \theta} > 0$
As $\theta$ increases, $\Delta$ increases $⇒$ when cos $\theta$ decreases, $\Delta$ increases
$∴ {\Delta}_{\text{min}}\text{occurs at cos}\theta = 1 / 2\text{,}\text{Therefore}{\Delta}_{2} = 2\sqrt{3} \cdot \dfrac{{\left(1 - 1 / 4\right)}^{3 / 2}}{1 / 2} = 4\sqrt{3} \cdot \dfrac{3\sqrt{3}}{8} = \dfrac{36}{8}$
${\Delta}_{\text{max}} \text{occurs at cos} \theta = 1 / 4 \text{,} \text{Therefore} {\Delta}_{1} = 2 \sqrt{3} \cdot \dfrac{{\left(1 - 1 / 1 6\right)}^{3 / 2}}{1 / 4} = 8 \sqrt{3} \cdot \dfrac{1 5 \text{.} \sqrt{1 5}}{\text{4.4.4}} = \dfrac{2 \sqrt{3} \text{.} 1 5 \text{.} \sqrt{3} \sqrt{5}}{1 6}$
$⇒ {\Delta}_{1} = \dfrac{4 5}{8} \sqrt{5}$
Now $\dfrac{8}{\sqrt{5}} {\Delta}_{1} - 8 {\Delta}_{2} = 4 5 - 3 6 = 9$