KVPY SX SB Exam 2018 Question Paper

$\left({\mathit{x}}^2∕9\right)+\left({\mathit{y}}^2∕4\right)=1$
Using parametric coordinate of the ellipse $\left(\mathit{a}\cos \mathit{\theta },\mathit{b}\sin \mathit{\theta }\right)=\left(3\cos \mathit{\theta },2\sin \mathit{\theta }\right)$

Equation of tangent at point $\left(3\cos \mathit{\theta },2\sin \mathit{\theta }\right)$ is $\mathit{T}=0\left\{\mathit{T}=\mathit{x}{\mathit{x}}_1∕{\mathit{a}}^2+\mathit{y}{\mathit{y}}_1∕{\mathit{b}}^2=1\right\}$
$\mathit{P}\mathit{Q}:\left[3\mathit{x}\cos \mathit{\theta }∕9\right]+\left[2\mathit{y}\sin \mathit{\theta }∕4\right]=1$
$\mathit{P}\mathit{Q}:\left[\mathit{x}\cos \mathit{\theta }∕3\right]+\left[\mathit{y}\sin \mathit{\theta }∕2\right]=1$
At $\mathit{x}=0\Rightarrow \mathit{y}=2\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }$
At $\mathit{y}=0\Rightarrow \mathit{x}=3\sec \mathit{\theta }$
Area of quadrilateral $\mathit{A}\left(\mathit{\theta }\right)=4\times$ Area of $△\mathit{O}\mathit{P}\mathit{Q}$
$=4\times \left(1∕2\right)\times$ Base (OP) $\times$ Height (OQ)
$=4\times \left(1∕2\right)\times 3\sec \mathit{\theta }.2\mathrm{cosec}\mathit{\theta }$
$=12∕\left[\sin \mathit{\theta }\cos \mathit{\theta }\right]$
$=24∕\sin 2\mathit{\theta }$
Minimum $\mathit{A}\left(\mathit{\theta }\right)=24∕{\left(\sin 2\mathit{\theta }\right)}_{\max }=24∕1=24$