Let $$x(t) = 2\sqrt{2}\cos t\sqrt{\sin 2t}$$ and $$y(t) = 2\sqrt{2}\sin t\sqrt{\sin 2t}$$, $$t \in (0, \frac{\pi}{2})$$. Then $$\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}$$ at $$t = \frac{\pi}{4}$$ is equal to

We are given the parametric curve $$x(t) = 2\sqrt{2}\cos t\sqrt{\sin 2t}$$ and $$y(t) = 2\sqrt{2}\sin t\sqrt{\sin 2t}$$ for $$t \in (0, \frac{\pi}{2})$$, and we need to evaluate $$\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}$$ at $$t = \frac{\pi}{4}$$.

We begin by finding $$\frac{dx}{dt}$$. Writing $$x = 2\sqrt{2}\cos t\,(\sin 2t)^{1/2}$$ and applying the product rule:

$$\frac{dx}{dt} = 2\sqrt{2}\left[-\sin t\,(\sin 2t)^{1/2} + \cos t \cdot \frac{2\cos 2t}{2(\sin 2t)^{1/2}}\right]$$

Combining terms over a common denominator $$(\sin 2t)^{1/2}$$:

$$\frac{dx}{dt} = \frac{2\sqrt{2}}{(\sin 2t)^{1/2}}\bigl[-\sin t\,\sin 2t + \cos t\,\cos 2t\bigr]$$

The expression in brackets equals $$\cos(t + 2t) = \cos 3t$$ (by the cosine addition formula). So $$\frac{dx}{dt} = \frac{2\sqrt{2}\cos 3t}{\sqrt{\sin 2t}}$$.

By the same approach for $$y = 2\sqrt{2}\sin t\,(\sin 2t)^{1/2}$$:

$$\frac{dy}{dt} = \frac{2\sqrt{2}}{(\sin 2t)^{1/2}}\bigl[\cos t\,\sin 2t + \sin t\,\cos 2t\bigr] = \frac{2\sqrt{2}\sin 3t}{\sqrt{\sin 2t}}$$

(using $$\cos t\,\sin 2t + \sin t\,\cos 2t = \sin(t + 2t) = \sin 3t$$).

At $$t = \frac{\pi}{4}$$, we have $$\sin 2t = 1$$, $$\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$, and $$\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$$. Substituting:

$$\frac{dx}{dt}\bigg|_{t=\pi/4} = 2\sqrt{2}\left(-\frac{\sqrt{2}}{2}\right) = -2, \qquad \frac{dy}{dt}\bigg|_{t=\pi/4} = 2\sqrt{2}\left(\frac{\sqrt{2}}{2}\right) = 2$$

So $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{-2} = -1$$.

Next, we need $$\frac{d^2y}{dx^2}$$. Using the formula $$\frac{d^2y}{dx^2} = \frac{\dot{x}\,\ddot{y} - \dot{y}\,\ddot{x}}{\dot{x}^3}$$, we must find the second derivatives $$\ddot{x}$$ and $$\ddot{y}$$ at $$t = \frac{\pi}{4}$$.

Differentiating $$\dot{x} = 2\sqrt{2}\cos 3t\,(\sin 2t)^{-1/2}$$ by the product rule:

$$\ddot{x} = 2\sqrt{2}\left[-3\sin 3t\,(\sin 2t)^{-1/2} - \cos 3t\,\frac{\cos 2t}{(\sin 2t)^{3/2}}\right]$$

At $$t = \frac{\pi}{4}$$, $$\cos 2t = \cos\frac{\pi}{2} = 0$$ and $$\sin 2t = 1$$, so the second term vanishes entirely. This gives:

$$\ddot{x}\big|_{t=\pi/4} = 2\sqrt{2}\left(-3 \cdot \frac{\sqrt{2}}{2}\right) = -6$$

Similarly, differentiating $$\dot{y} = 2\sqrt{2}\sin 3t\,(\sin 2t)^{-1/2}$$:

$$\ddot{y} = 2\sqrt{2}\left[3\cos 3t\,(\sin 2t)^{-1/2} - \sin 3t\,\frac{\cos 2t}{(\sin 2t)^{3/2}}\right]$$

At $$t = \frac{\pi}{4}$$, the second term again vanishes, so:

$$\ddot{y}\big|_{t=\pi/4} = 2\sqrt{2}\left(3 \cdot \left(-\frac{\sqrt{2}}{2}\right)\right) = -6$$

Now substituting into the second derivative formula:

$$\frac{d^2y}{dx^2} = \frac{(-2)(-6) - (2)(-6)}{(-2)^3} = \frac{12 + 12}{-8} = \frac{24}{-8} = -3$$

Finally, the required expression is:

$$\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}} = \frac{1 + (-1)^2}{-3} = \frac{2}{-3} = -\frac{2}{3}$$

Hence, the correct answer is Option 4.