1. **Stating the problem:** We are given the implicit equation $$ (x^2 + y^2)^2 = xy $$ and asked to analyze or understand it. 2. **Understanding the equation:** This is an implicit relation between $x$ and $y$, not a function $y=f(x)$ explicitly. It involves both $x$ and $y$ in a nonlinear way. 3. **Rewrite the equation:** The equation is $$ (x^2 + y^2)^2 = xy $$ which means the square of the sum of squares of $x$ and $y$ equals the product $xy$. 4. **Important observations:** - Since the left side is always nonnegative (a square), the right side $xy$ must also be nonnegative. - This implies $xy \geq 0$, so $x$ and $y$ must be both nonnegative or both nonpositive. 5. **Check for trivial solutions:** - If $x=0$ or $y=0$, then right side is zero. - Left side is $(x^2 + y^2)^2$, which is zero only if $x=0$ and $y=0$. - So $(0,0)$ is a solution. 6. **Try to express $y$ in terms of $x$ (if possible):** - Let $r^2 = x^2 + y^2$. - Then equation is $$ r^4 = xy $$ - But $r^2 = x^2 + y^2$, so $r^4 = (x^2 + y^2)^2$. 7. **Parametric approach:** - Let $y = tx$ (assuming $x \neq 0$), then $$ (x^2 + (tx)^2)^2 = x (tx) $$ $$ (x^2 + t^2 x^2)^2 = t x^2 $$ $$ (x^2 (1 + t^2))^2 = t x^2 $$ $$ x^4 (1 + t^2)^2 = t x^2 $$ 8. **Divide both sides by $x^2$ (assuming $x \neq 0$):** $$ x^2 (1 + t^2)^2 = t $$ 9. **Solve for $x^2$:** $$ x^2 = \frac{t}{(1 + t^2)^2} $$ 10. **Find $y$:** $$ y = t x = t \sqrt{\frac{t}{(1 + t^2)^2}} = \frac{t \sqrt{t}}{(1 + t^2)} $$ 11. **Domain considerations:** - For $x^2$ to be real and nonnegative, $t \geq 0$. - So $t \geq 0$. 12. **Summary:** - The curve can be parametrized as $$ x = \pm \sqrt{\frac{t}{(1 + t^2)^2}} $$ $$ y = t x $$ for $t \geq 0$. 13. **Final remarks:** - The implicit curve defined by $(x^2 + y^2)^2 = xy$ is symmetric about the origin. - The only trivial solution is $(0,0)$. - The parametric form helps to understand the shape and domain of the curve. **Answer:** The implicit curve $(x^2 + y^2)^2 = xy$ can be parametrized by $$ x = \pm \sqrt{\frac{t}{(1 + t^2)^2}}, \quad y = t x, \quad t \geq 0 $$ with the trivial solution at $(0,0)$.