1. The problem asks to show that for the curve $y = \sqrt{x} \sin 2x$ on $0 \leq x \leq \frac{1}{2} \pi$, the maximum point $M$ at $x = a$ satisfies $\tan 2a = -4a$. 2. To find the maximum, we differentiate $y$ with respect to $x$ and set the derivative equal to zero. 3. Using the product rule, $y = x^{1/2} \sin 2x$, so $$\frac{dy}{dx} = \frac{1}{2} x^{-1/2} \sin 2x + \sqrt{x} \cdot 2 \cos 2x.$$ 4. Simplify: $$\frac{dy}{dx} = \frac{\sin 2x}{2 \sqrt{x}} + 2 \sqrt{x} \cos 2x.$$ 5. Set $\frac{dy}{dx} = 0$ for the maximum: $$\frac{\sin 2a}{2 \sqrt{a}} + 2 \sqrt{a} \cos 2a = 0.$$ 6. Multiply both sides by $2 \sqrt{a}$ to clear the denominator: $$\sin 2a + 4a \cos 2a = 0.$$ 7. Rearranged: $$\sin 2a = -4a \cos 2a.$$ 8. Divide both sides by $\cos 2a$ (assuming $\cos 2a \neq 0$): $$\tan 2a = -4a.$$ This completes the proof. Final answer: $\boxed{\tan 2a = -4a}$.