Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 39

Find the points where the tangent is horizontal on the lemniscate having a function of $2(x^2 + y^2)^2 = 25(x^2 - y^2)$

Taking the derivative of both sides of the equation implicitly we have


$
\begin{equation}
\begin{aligned}

& 4(x^2 + y^2) (2x + 2y y') = 25(2x - 2y y')
\\
\\
& 8x^3 + 8x^2yy' + 8xy^2 + 8y^3y' = 50x - 50yy'
\\
\\
& 8x^2yy' + 8y^3 y' + 50yy' = 50x - 8x^3 - 8xy^2
\\
\\
& y' = \frac{x [50 - 8x^2 - 8y^2]}{8x^2 y + 8y^3 + 50y}

\end{aligned}
\end{equation}
$


The equation of the horizontal tangent can be determined where $y' = 0$ so..


$
\begin{equation}
\begin{aligned}

& 0 = x [50 - 8x^2 - 8y^2]
\\
\\
& 8x^2 + 8y^2 = 50
\\
\\
& x^2 + y^2 = \frac{25}{4} \qquad \text{Equation 1}


\end{aligned}
\end{equation}
$


From the equation of the lemniscate,


$
\begin{equation}
\begin{aligned}

2\left( \frac{25}{4} \right) ^2 =& 25 (x^2 - y^2)
\\
\\
x^2 - y^2 =& \frac{25}{8 } \qquad \text{Equation 2}

\end{aligned}
\end{equation}
$


Use Equations 1 and 2 to locate the points where the horizontal tangent is .


$
\begin{equation}
\begin{aligned}

x^2 + y^2 =& \frac{25}{4}
\\
\\
x^2 - y^2 =& \frac{25}{8}

\end{aligned}
\end{equation}
$


Solving for $x$ and $y$ simultaneously, we have..

$\displaystyle x = \pm \frac{5 \sqrt{3}}{4}$ and $\displaystyle y = \pm \frac{5}{4}$

Therefore, we have four points at $\displaystyle \left( \pm \frac{5 \sqrt{3}}{4} , \pm \frac{5}{4} \right)$