Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 29

Find an equation of the tangent to the curve $2(x^2 + y^2)^2 = 25(x^2 - y^2)$ at the point $(3,1)$ using implicit differentiation if $y ' = m$ then


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} 2(x^2 + y^2)^2 =& \frac{d}{dx} 25(x^2 - y^2)
\\
\\
2 \frac{d}{dx} (x^2 + y^2)^2 =& 25 \frac{d}{dx} (x^2 - y^2)
\\
\\
(2)(2) (x^2 + y^2) \frac{d}{dx} (x^2 + y^2) =& (25)\left( 2x - 2y \frac{dy}{dx} \right)
\\
\\
4(x^2 + y^2) \left( 2x + 2y \frac{dy}{dx} \right) =& 50 x - 50y \frac{dy}{dx}
\\
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8x^3 + 8x^2y \frac{dy}{dx} + 8xy^2 + 8y^3 \frac{dy}{dx} =& 50x - 50y \frac{dy}{dx}
\\
\\
8x^2yy' + 8y^3y' + 50yy' =& 50x - 8x^3 - 8xy^2
\\
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y'(8x^2y + 8y^3 + 50y) =& 50x - 8x^3 - 8xy^2
\\
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\frac{y'(\cancel{8x^2y + 8y^3 + 50y})}{\cancel{8x^2y + 8y^3 + 50y}} =& \frac{50x - 8x^3 - 8xy^2}{8x^2y + 8y^3 + 50y}
\\
\\
y =& \frac{50x - 8x^3 - 8xy^2}{8x^2y + 8y^3 + 50y}
\end{aligned}
\end{equation}
$



For $x = 3$ and $y = 1$, we obtain


$
\begin{equation}
\begin{aligned}

y' = m =& \frac{50(3) - 8(3)^3 - 8(3)(1)^2}{8(3)^2 (1) + 8 (1)^3 + 50(1)}
\\
\\
m =& \frac{-90}{130}
\\
\\
m =& \frac{-9}{13}

\end{aligned}
\end{equation}
$


Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
\\
\\
y - 1 =& \frac{-9}{13} (x - 3)
\\
\\
y =& \frac{-9x + 27}{13} + 1
\\
\\
y =& \frac{-9x + 27 + 13}{13}
\\
\\
y =& \frac{-9x + 40}{13} \qquad \text{Equation of the tangent line at $(3, 1)$}

\end{aligned}
\end{equation}
$