1. **State the problem:** We need to trace the curve given by the equation $$y^2(a^2 + x^2) = x^2(a^2 - x^2)$$ where $a$ is a constant. 2. **Rewrite the equation:** The equation relates $y^2$ and $x^2$. We want to express $y^2$ in terms of $x^2$ to understand the curve. 3. **Isolate $y^2$:** $$y^2 = \frac{x^2(a^2 - x^2)}{a^2 + x^2}$$ 4. **Analyze the domain:** For $y^2$ to be real and non-negative, the numerator and denominator must be such that the fraction is non-negative. - Denominator $a^2 + x^2 > 0$ for all real $x$. - Numerator $x^2(a^2 - x^2) \geq 0$. Since $x^2 \geq 0$, the sign depends on $(a^2 - x^2)$. 5. **Domain for $x$:** $$a^2 - x^2 \geq 0 \implies |x| \leq a$$ 6. **Express $y$ explicitly:** $$y = \pm \sqrt{\frac{x^2(a^2 - x^2)}{a^2 + x^2}}$$ 7. **Check intercepts:** - At $x=0$, $$y^2 = 0 \implies y=0$$ - At $y=0$, $$0 = \frac{x^2(a^2 - x^2)}{a^2 + x^2} \implies x=0 \text{ or } x=\pm a$$ 8. **Behavior at boundaries:** - At $x=\pm a$, $$y^2 = 0 \implies y=0$$ 9. **Summary:** The curve is symmetric about both axes, defined for $|x| \leq a$, with $y=0$ at $x=0$ and $x=\pm a$. The shape resembles a closed curve between $-a$ and $a$ on the $x$-axis. **Final answer:** $$y = \pm \sqrt{\frac{x^2(a^2 - x^2)}{a^2 + x^2}}$$