consider the curve described by the equation (x² + y²)² = 4(x² - y²). (a) 6pts. use the method of implicit…

Explanation:

Step1: Differentiate left - hand side

Differentiate $(x^{2}+y^{2})^{2}$ with respect to $x$ using the chain - rule. Let $u = x^{2}+y^{2}$, so $(x^{2}+y^{2})^{2}=u^{2}$. Then $\frac{d}{dx}(u^{2}) = 2u\frac{du}{dx}$, and $\frac{du}{dx}=2x + 2y y'$. So $\frac{d}{dx}(x^{2}+y^{2})^{2}=2(x^{2}+y^{2})(2x + 2y y')$.

Step2: Differentiate right - hand side

Differentiate $4(x^{2}-y^{2})$ with respect to $x$. Using the sum/difference rule and the chain - rule, $\frac{d}{dx}(4(x^{2}-y^{2}))=4(2x-2y y')$.

Step3: Set derivatives equal

Set $2(x^{2}+y^{2})(2x + 2y y')=4(2x-2y y')$. Expand the left - hand side: $4x(x^{2}+y^{2})+4y(x^{2}+y^{2})y'=8x - 8y y'$. Collect terms with $y'$: $4y(x^{2}+y^{2})y'+8y y'=8x - 4x(x^{2}+y^{2})$. Factor out $y'$: $y'(4y(x^{2}+y^{2}) + 8y)=8x - 4x(x^{2}+y^{2})$. Solve for $y'$: $y'=\frac{8x - 4x(x^{2}+y^{2})}{4y(x^{2}+y^{2})+8y}=\frac{2x - x(x^{2}+y^{2})}{y(x^{2}+y^{2}) + 2y}$.

Step4: Find horizontal tangent points

A horizontal tangent line occurs when $y' = 0$. So $2x - x(x^{2}+y^{2})=0$ and $y(x^{2}+y^{2})+2y\neq0$. Factor the equation $2x - x(x^{2}+y^{2})=0$: $x(2-(x^{2}+y^{2}))=0$. Case 1: $x = 0$. Substitute $x = 0$ into the original equation $(x^{2}+y^{2})^{2}=4(x^{2}-y^{2})$, we get $(0 + y^{2})^{2}=4(0 - y^{2})$, i.e., $y^{4}+4y^{2}=0$. Let $t = y^{2}$, then $t^{2}+4t = 0$, $t(t + 4)=0$. Since $t=y^{2}\geq0$, $t = 0$, so $y = 0$. Case 2: $x^{2}+y^{2}=2$. Substitute $x^{2}+y^{2}=2$ into the original equation $(x^{2}+y^{2})^{2}=4(x^{2}-y^{2})$, we have $4 = 4(x^{2}-y^{2})$, so $x^{2}-y^{2}=1$. Solve the system $\begin{cases}x^{2}+y^{2}=2\x^{2}-y^{2}=1\end{cases}$ by adding the two equations: $2x^{2}=3$, $x=\pm\frac{\sqrt{6}}{2}$. Substitute $x=\pm\frac{\sqrt{6}}{2}$ into $x^{2}+y^{2}=2$, we get $y=\pm\frac{\sqrt{2}}{2}$.

Answer:

(a) $y'=\frac{2x - x(x^{2}+y^{2})}{y(x^{2}+y^{2}) + 2y}$ (b) $(\frac{\sqrt{6}}{2},\frac{\sqrt{2}}{2}),(\frac{\sqrt{6}}{2},-\frac{\sqrt{2}}{2}),(-\frac{\sqrt{6}}{2},\frac{\sqrt{2}}{2}),(-\frac{\sqrt{6}}{2},-\frac{\sqrt{2}}{2})$