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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 $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$
$6$
$7$
$8$
$9$
Solution
Point of intersection of tangents at $P$ and $Q$ is $R(2 \sec \theta, 0)$
Area of $\triangle PQR =\frac{1}{2} \cdot 2 \sqrt{3} \sin \theta \cdot(2 \sec \theta-2 \cos \theta)$
$\Rightarrow \quad \Delta=2 \sqrt{3} \cdot \frac{\sin ^3 \theta}{\cos \theta} ;$ where $\cos \theta \in\left[\frac{1}{4}, \frac{1}{2}\right]$
$Image$
Now $\quad \frac{ d \Delta}{ d \theta}=\frac{\left.2 \sqrt{3} \mid \cos \theta \cdot 3 \sin ^2 \theta \cos \theta-\sin ^3 \theta(-\sin \theta)\right]}{\cos ^2 \theta}>0$
As $\theta$ increases, $\Delta$ increases $\Rightarrow$ when $\cos \theta$ decreases, $\Delta$ increases
$\therefore \quad \Delta_{\min .} \text { occurs at } \cos \theta=1 / 2 \text {, Therefore } \Delta_2=2$ $\sqrt{3} \cdot \frac{(1-1 / 4)^{3 / 2}}{1 / 2}=4 \sqrt{3} \cdot \frac{3 \sqrt{3}}{8}=\frac{36}{8} $
$\Delta_{\max } \text { occurs at } \cos \theta=1 / 4 \text {, Therefore } \Delta_1=2 \sqrt{3} \cdot \frac{(1-1 / 16)^{3 / 2}}{1 / 4}=8 \sqrt{3} \cdot \frac{15 \cdot \sqrt{15}}{4 \cdot 4.4}=\frac{2 \sqrt{3} \cdot 15 \cdot \sqrt{3} \sqrt{5}}{16} $
$\Rightarrow \quad \Delta_1=\frac{45}{8} \sqrt{5}$
Now $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=45-36=9$
