Let $A\left(\frac{3}{\sqrt{a}}, \sqrt{a}\right), a0$, be a fixed point in the $x y$-plane. The imag

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Let $A\left(\frac{3}{\sqrt{a}}, \sqrt{a}ight), a>0$, be a fixed point in the $x y$-plane. The image of $A$ in $y$-axis be $B$ and the image of $B$ in $x$-axis be $C$. If $D(3 \cos 	heta, a \sin 	heta)$ is a point in the fourth quadrant such that the maximum area of $	riangle A C D$ is 2 square units, then $a$ is equal to

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(8) : We have, $A=\left(\frac{3}{\sqrt{a}}, \sqrt{a}ight), a>0$ Image of $A$ in $y$-axis is $\left(\frac{-3}{\sqrt{a}}, \sqrt{a}ight)$\ So, coordinates of $B=\left(\frac{-3}{\sqrt{a}}, \sqrt{a}ight)$\ Similarly, coordinates of $C=\left(\frac{-3}{\sqrt{a}},-\sqrt{a}ight)$\ Now, maximum area of $	riangle A C D=12$\ $\Rightarrow \max \cdot\left[\frac{1}{2}\left|\begin{array}{ccc}3 / \sqrt{a} & \sqrt{a} & 1 \ -3 / \sqrt{a} & -\sqrt{a} & 1 \ 3 \cos 	heta & a \sin 	heta & 1\end{array}ight|ight]= \pm 12$\ Applying $R_{1} ightarrow R_{1}+R_{2}$\ $\Rightarrow \max \cdot\left[\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 2 \ -3 / \sqrt{a} & -\sqrt{a} & 1 \ 3 \cos 	heta & a \sin 	heta & 1\end{array}ight|ight]= \pm 12$\ $\Rightarrow \quad \max \cdot[(-3 \sqrt{a} \sin 	heta+3 \sqrt{a} \cos 	heta)]= \pm 12$\ $\Rightarrow \quad \max \cdot[\sqrt{a}(\cos 	heta-\sin 	heta)]= \pm 4$\ Maximum value of\ $\cos 	heta-\sin 	heta=\sqrt{1^{2}+1^{2}}=\sqrt{2}$\ $	herefore \quad \sqrt{a} \sqrt{2}= \pm 4 \Rightarrow 2 a=16 \Rightarrow a=8$\