Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 30

Find an equation of the tangent to the curve $y^2 (y^2 - 4) = x^2 (x^2 - 5)$ at the point $( 0,-2)$ using implicit differentiation if $y' = m$ then

$\displaystyle \frac{d}{dx} [y^2 (y^2 - 4)] = \frac{d}{dx} [x^2(x^2 - 5)]$


$
\begin{equation}
\begin{aligned}

(y^2) \frac{d}{dx} (y^2 - 4) + (y^2 - 4) \frac{d}{dx} (y^2) =& (x^2) \frac{d}{dx} (x^2 - 5) + (x^2 - 5) \frac{d}{dx} (x^2)
\
\\
(y^2) \left( 2y \frac{dy}{dx} - 0 \right) + (y^2 - 4) (2y) \frac{dy}{dx} =& (x^2) (2x - 0 ) + (x^2 - 5)(2x)
\\
\\
2y^3 \frac{dy}{dx} + 2y^3 \frac{dy}{dx} - 8y \frac{dy}{dx} =& 2x^3 + 2x^3 - 10x
\\
\\
4y^3 y' - 8yy' =& 4x^3 - 10x
\\
\\
y'(4y^3 - 8y) =& 4x^3 - 10x
\\
\\
\frac{y' (\cancel{4y^3 - 8y})}{\cancel{4y^3 - 8y}} =& \frac{4x^3 - 10x}{4y^3 - 8y}
\\
\\
y' =& \frac{4x^3 - 10x}{4y^3 - 8y}
\\
\\
y' =& \frac{2(2x^3 - 5x)}{2(2y^3 - 4y)}
\\
\\
y' =& \frac{2x^3 - 5x}{2y^3 - 4y}

\end{aligned}
\end{equation}
$


For $x = 0$ and $y = -2$


$
\begin{equation}
\begin{aligned}

y' = m =& \frac{2(0)^3 - 5(0)}{2(-2)^3 - 4 (-2)}
\\
\\
m =& \frac{0}{-8}
\\
\\
m =& 0

\end{aligned}
\end{equation}
$


Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
\\
\\
y - (-2) =& 0(x - 0)
\\
\\
y + 2 =& 0
\\
\\
y =& -2 \qquad \text{Equation of the tangent line at $(0, -2)$}

\end{aligned}
\end{equation}
$