Question

Let $A \left(\frac{3}{\sqrt{ a }}, \sqrt{ a }\right) a >0$, 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 $D (3 \cos \theta$, a $\sin \theta)$ is a point in the fourth quadrant such that the maximum area of $\triangle ACD$ is $12$ square units, then a is equal to __

Answer 1

$ A =\left(\frac{3}{\sqrt{ a }}, \sqrt{ a }\right) $
$ B =\left(\frac{-3}{\sqrt{ a }}, \sqrt{ a }\right) $
$ C =\left(-\frac{3}{\sqrt{ a }},-\sqrt{ a }\right)$
Area of $ACD$
$\frac{1}{2}\begin{vmatrix}\frac{3}{\sqrt{a}} & \sqrt{a} \\ -\frac{3}{\sqrt{a}} & -\sqrt{a} \\ 3 \cos \theta & a \sin \theta \\ \frac{3}{\sqrt{a}} & \sqrt{a}\end{vmatrix}$
$\frac{1}{2} 6 \sqrt{a}(\cos \theta-\sin \theta)$
$3 \sqrt{a}(\cos \theta-\sin \theta)$
$\max$ values of function is $3 \sqrt{ a } \sqrt{2}$
$3 \sqrt{ a } \sqrt{2}=12$
$2 a =16 $
$ a =8$