Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 40

By using implicit differentiation, show that the tangent to the ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(x_0, y_0)$ is $\displaystyle \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1$

Taking the derivative of the ellipse implicitly we get


$
\begin{equation}
\begin{aligned}

\frac{1}{a^2} (2x) + \frac{1}{b^2} \left( 2y \frac{dy}{dx} \right) =& 0
\\
\\
\frac{dy}{dx} =& \frac{xb^2}{ya^2}

\end{aligned}
\end{equation}
$



Using Point Slope Form @ $(x_0, y_0)$


$
\begin{equation}
\begin{aligned}

y - y_0 =& m(x - x_0)
\\
\\
y - y_0 =& \frac{xb^2}{ya^2} (x - x_0)

\end{aligned}
\end{equation}
$


Multiplying $\displaystyle \frac{y}{b^2}$ on both sides of the equation we have..


$
\begin{equation}
\begin{aligned}

\frac{y}{b^2} (y - y_0) =& \frac{x}{a^2} (x - x_0)
\\
\\
\frac{y^2}{b^2} - \frac{yy_0}{b^2} =& \frac{x^2}{a^2} - \frac{xx_0}{a^2}
\\
\\
\frac{xx_0}{a^2} - \frac{yy_0}{b^2} =& \frac{x^2}{a^2} - \frac{y^2}{b^2}

\end{aligned}
\end{equation}
$


From the given equation, we know that $\displaystyle \left( \frac{x^2}{a^2} - \frac{y^2}{b^2} \right) = 1$ so..

$\displaystyle \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$\

Hence, the equation of the tangent line at point $(x_0, y_0)$