1. **State the problem:** Solve the equation $r^2 \sin 2\theta = 2a^2$ for $r$ in terms of $\theta$ and $a$. 2. **Recall the formula and rules:** The equation involves polar coordinates where $r$ is the radius and $\theta$ is the angle. 3. **Isolate $r^2$:** $$r^2 = \frac{2a^2}{\sin 2\theta}$$ 4. **Take the square root of both sides:** $$r = \pm \sqrt{\frac{2a^2}{\sin 2\theta}}$$ 5. **Simplify the expression:** $$r = \pm a \sqrt{\frac{2}{\sin 2\theta}}$$ 6. **Important note:** $\sin 2\theta$ must not be zero to avoid division by zero. **Final answer:** $$r = \pm a \sqrt{\frac{2}{\sin 2\theta}}$$