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 $$\triangle ACD$$ is $$12$$ square units, then $$a$$ is equal to ______

Given $$A = \left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$$, $$B$$ is the image of $$A$$ in the y-axis, and $$C$$ is the image of $$B$$ in the x-axis.

Find coordinates of B and C.

$$B = \left(-\frac{3}{\sqrt{a}}, \sqrt{a}\right)$$ (reflection of A in y-axis)

$$C = \left(-\frac{3}{\sqrt{a}}, -\sqrt{a}\right)$$ (reflection of B in x-axis)

Set up the area of triangle ACD.

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

Area $$= \frac{1}{2}|x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C)|$$

$$= \frac{1}{2}\left|\frac{3}{\sqrt{a}}(-\sqrt{a} - a\sin\theta) + \left(-\frac{3}{\sqrt{a}}\right)(a\sin\theta - \sqrt{a}) + 3\cos\theta(\sqrt{a} + \sqrt{a})\right|$$

$$= \frac{1}{2}\left|(-3 - 3\sqrt{a}\sin\theta) + (-3\sqrt{a}\sin\theta + 3) + 6\sqrt{a}\cos\theta\right|$$

$$= \frac{1}{2}\left|-6\sqrt{a}\sin\theta + 6\sqrt{a}\cos\theta\right|$$

$$= 3\sqrt{a}|\cos\theta - \sin\theta|$$

Maximize the area.

The maximum of $$|\cos\theta - \sin\theta| = \sqrt{2}$$.

Maximum area $$= 3\sqrt{a} \cdot \sqrt{2} = 12$$

$$\sqrt{a} = \frac{12}{3\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$

$$a = 8$$

Answer: 8