Question

Let $A\left(\frac{3}{\sqrt{a}},\sqrt{a}\right)a>0$, be a fixed point in the $xy$-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 $△ACD$ is $12$ square units, then a is equal to __

Answer 1

Answer: 8

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