Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 41

Determine the equation of the tangent line to the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(x_0, y_0)$

Taking the derivative of the curve implicitly we have...


$\begin{equation} \begin{aligned} \frac{1}{a^2} (2x) - \frac{1}{b^2} \left( 2y \frac{dy}{dx} \right) =& 0 \\ \\ \frac{dy}{dx} =& \frac{xb^2}{ya^2} \end{aligned} \end{equation}$


Using Point Slope Form


$\begin{equation} \begin{aligned} y - y_0 =& m(x - x_0) \\ \\ y - y_0 =& \frac{xb^2}{ya^2} (x - x_0) \end{aligned} \end{equation}$


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


$\begin{equation} \begin{aligned} \frac{y}{b^2} (y - y_0) =& \frac{x}{a^2} (x - x_0) \\ \\ \frac{y^2}{b^2} - \frac{y y_0}{b^2} =& \frac{x^2}{a^2} - \frac{xx_0}{a^2} \\ \\ \frac{xx_0}{a^2} - \frac{yy_0}{b^2} =& \frac{x^2}{a^2} - \frac{y^2}{b^2} \end{aligned} \end{equation}$


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 $(x_0, y_0)$