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Question

Point of intersection of tangents at $P$ and $Q$ is $R(2 \sec 	heta, 0)$

Area of $	riangle PQR =\frac{1}{2} \cdot 2 \sqrt{3} \sin 	heta \cdot(2

Vedclass · Accepted Answer

$9$

Vedclass · Answer

Point of intersection of tangents at $P$ and $Q$ is $R(2 \sec 	heta, 0)$

Area of $	riangle PQR =\frac{1}{2} \cdot 2 \sqrt{3} \sin 	heta \cdot(2 \sec 	heta-2 \cos 	heta)$

$\Rightarrow \quad \Delta=2 \sqrt{3} \cdot \frac{\sin ^3 	heta}{\cos 	heta} ;$ where $\cos 	heta \in\left[\frac{1}{4}, \frac{1}{2}ight]$

$Image$

Now $\quad \frac{ d \Delta}{ d 	heta}=\frac{\left.2 \sqrt{3} \mid \cos 	heta \cdot 3 \sin ^2 	heta \cos 	heta-\sin ^3 	heta(-\sin 	heta)ight]}{\cos ^2 	heta}>0$

As $	heta$ increases, $\Delta$ increases $\Rightarrow$ when $\cos 	heta$ decreases, $\Delta$ increases

$	herefore \quad \Delta_{\min .} 	ext { occurs at } \cos 	heta=1 / 2 	ext {, 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 } 	ext { occurs at } \cos 	heta=1 / 4 	ext {, 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 $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=45-36=9$

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