1. Coordinates of the points

Given $A\left(\frac{3}{\sqrt a},\sqrt a\right), \quad a>0.$

• Image of $A$ in the $y$-axis: $B\left(-\frac{3}{\sqrt a},\sqrt a\right).$
• Image of $B$ in the $x$-axis: $C\left(-\frac{3}{\sqrt a},-\sqrt a\right).$

Also, $D(3\cos\theta, a\sin\theta).$ Since $D$ is in the fourth quadrant, $3\cos\theta>0,\quad a\sin\theta<0 \implies \cos\theta>0,\ \sin\theta<0.$

2. Area of $\triangle ACD$

Using coordinates:

• $A\left(\frac{3}{\sqrt a},\sqrt a\right)$
• $C\left(-\frac{3}{\sqrt a},-\sqrt a\right)$
• $D(3\cos\theta,a\sin\theta)$

Area formula: $[\triangle ACD]=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$

Substituting, $[\triangle ACD]=\frac12\left|\frac{3}{\sqrt a}(-\sqrt a-a\sin\theta)+\left(-\frac{3}{\sqrt a}\right)(a\sin\theta-\sqrt a)+3\cos\theta(\sqrt a-(-\sqrt a))\right|.$

Now simplify term by term:

$\frac{3}{\sqrt a}(-\sqrt a-a\sin\theta)=-3-3\sqrt a\sin\theta,$ $\left(-\frac{3}{\sqrt a}\right)(a\sin\theta-\sqrt a)=-3\sqrt a\sin\theta+3,$ $3\cos\theta(2\sqrt a)=6\sqrt a\cos\theta.$

So, $[\triangle ACD]=\frac12\left|-6\sqrt a\sin\theta+6\sqrt a\cos\theta\right|$ $=3\sqrt a\,|\cos\theta-\sin\theta|.$

3. Maximize the area

Since $D$ is in the fourth quadrant, $\cos\theta>0$ and $\sin\theta<0$, hence $\cos\theta-\sin\theta>0.$ Therefore, $[\triangle ACD]=3\sqrt a(\cos\theta-\sin\theta).$

We know $(\cos\theta-\sin\theta)^2=\cos^2\theta+\sin^2\theta-2\sin\theta\cos\theta=1-\sin2\theta.$ Its maximum value is $2$, so $\max(\cos\theta-\sin\theta)=\sqrt2.$

Hence maximum area is $[\triangle ACD]_{\max}=3\sqrt a\cdot\sqrt2=3\sqrt{2a}.$

Given this maximum area is $12$: $3\sqrt{2a}=12$ $\sqrt{2a}=4$ $2a=16$ $a=8.$

4. Comparison with stored answer

Derived answer: $a=8.$

Stored correct answer: $8$

They match.