find the points on the lemniscate where the tangent is horizontal. (order your answers from smallest to…

Explanation:

Step1: Differentiate implicitly

Differentiate $8(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})$ with respect to $x$. Using the chain - rule and product - rule, we have: [ \begin{align*} 16(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})&=25(2x-2y\frac{dy}{dx})\ 32x(x^{2}+y^{2})+32y(x^{2}+y^{2})\frac{dy}{dx}&=50x - 50y\frac{dy}{dx}\ 32y(x^{2}+y^{2})\frac{dy}{dx}+ 50y\frac{dy}{dx}&=50x - 32x(x^{2}+y^{2})\ \frac{dy}{dx}(32y(x^{2}+y^{2})+50y)&=50x - 32x(x^{2}+y^{2})\ \frac{dy}{dx}&=\frac{50x - 32x(x^{2}+y^{2})}{32y(x^{2}+y^{2})+50y} \end{align*} ]

Step2: Set $\frac{dy}{dx}=0$

For a horizontal tangent, $\frac{dy}{dx} = 0$. So, $50x-32x(x^{2}+y^{2})=0$. Factor out $2x$: $2x(25 - 16(x^{2}+y^{2}))=0$. Case 1: $x = 0$. Substitute $x = 0$ into the original equation $8(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})$: [ \begin{align*} 8y^{4}&=- 25y^{2}\ 8y^{4}+25y^{2}&=0\ y^{2}(8y^{2}+25)&=0 \end{align*} ] Since $8y^{2}+25>0$ for all real $y$, then $y = 0$. Case 2: $25-16(x^{2}+y^{2})=0$, so $x^{2}+y^{2}=\frac{25}{16}$. Substitute $y^{2}=\frac{25}{16}-x^{2}$ into the original equation $8(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})$: [ \begin{align*} 8\times(\frac{25}{16})^{2}&=25\left(x^{2}-\left(\frac{25}{16}-x^{2}\right)\right)\ 8\times\frac{625}{256}&=25\left(2x^{2}-\frac{25}{16}\right)\ \frac{625}{32}&=50x^{2}-\frac{625}{16}\ 50x^{2}&=\frac{625}{32}+\frac{625}{16}\ 50x^{2}&=\frac{625 + 1250}{32}\ 50x^{2}&=\frac{1875}{32}\ x^{2}&=\frac{1875}{32\times50}=\frac{75}{64}\ x&=\pm\frac{5\sqrt{3}}{8} \end{align*} ] When $x=\pm\frac{5\sqrt{3}}{8}$, $y^{2}=\frac{25}{16}-\frac{75}{64}=\frac{100 - 75}{64}=\frac{25}{64}$, so $y=\pm\frac{5}{8}$

Answer:

$(x,y)=\left(-\frac{5\sqrt{3}}{8},-\frac{5}{8}\right)$ $(x,y)=\left(-\frac{5\sqrt{3}}{8},\frac{5}{8}\right)$ $(x,y)=\left(\frac{5\sqrt{3}}{8},-\frac{5}{8}\right)$ $(x,y)=\left(\frac{5\sqrt{3}}{8},\frac{5}{8}\right)$