# Equation of the Tangent and Normal to a Hyperbola

The equations of the tangent and normal to the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ at the point $\left( {{x_1},{y_1}} \right)$ are $\frac{{{x_1}x}}{{{a^2}}} – \frac{{{y_1}y}}{{{b^2}}} = 1$ and ${a^2}{y_1}x + {b^2}{x_1}y – \left( {{a^2} + {b^2}} \right){x_1}{y_1} = 0$ respectively.

Consider that the standard equation of a hyperbola with vertex at origin $\left( {0,0} \right)$ can be written as
$\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Since the point $\left( {{x_1},{y_1}} \right)$ lies on the given hyperbola, it must satisfy equation (i). So we have

$\frac{{{x_1}^2}}{{{a^2}}} – \frac{{{y_1}^2}}{{{b^2}}} = 1\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Now differentiating equation (i) on both sides with respect to $x$, we have
$\begin{gathered} \frac{{2x}}{{{a^2}}} – \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0 \Rightarrow \frac{y}{{{b^2}}}\frac{{dy}}{{dx}} = \frac{x}{{{a^2}}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{{b^2}x}}{{{a^2}y}} \\ \end{gathered}$

If $m$ represents the slope of the tangent at the given point $\left( {{x_1},{y_1}} \right)$, then
$m = {\frac{{dy}}{{dx}}_{\left( {{x_1},{y_1}} \right)}} = \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}$

The equation of the tangent at the given point $\left( {{x_1},{y_1}} \right)$ is

$\begin{gathered} y – {y_1} = \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}\left( {x – {x_1}} \right) \\ \frac{{{y_1}}}{{{b^2}}}\left( {y – {y_1}} \right) = \frac{{{x_1}}}{{{a^2}}}\left( {x – {x_1}} \right) \\ \Rightarrow \frac{{{y_1}y}}{{{b^2}}} – \frac{{{y_1}^2}}{{{b^2}}} = \frac{{{x_1}x}}{{{a^2}}} – \frac{{{x_1}^2}}{{{a^2}}} \\ \Rightarrow \frac{{{x_1}x}}{{{a^2}}} – \frac{{{y_1}y}}{{{b^2}}} = \frac{{{x_1}^2}}{{{a^2}}} – \frac{{{y_1}^2}}{{{b^2}}} \\ \Rightarrow \boxed{\frac{{{x_1}x}}{{{a^2}}} – \frac{{{y_1}y}}{{{b^2}}} = 1} \\ \end{gathered}$

This is the equation of the tangent to the given hyperbola at $\left( {{x_1},{y_1}} \right)$.

The slope of the normal at $\left( {{x_1},{y_1}} \right)$ is $– \frac{1}{m} = – \left( {\frac{{{a^2}{x_1}}}{{{b^2}{y_1}}}} \right) = – \frac{{{a^2}{x_1}}}{{{b^2}{y_1}}}$

The equation of the normal at the point $\left( {{x_1},{y_1}} \right)$ is $y – {y_1} = – \frac{{{a^2}{x_1}}}{{{b^2}{y_1}}}\left( {x – {x_1}} \right)$
$\begin{gathered} \Rightarrow {a^2}{y_1}x + {b^2}{x_1}y – {a^2}{x_1}{y_1} – {b^2}{x_1}{y_1} = 0 \\ \Rightarrow \boxed{{a^2}{y_1}x + {b^2}{x_1}y – \left( {{a^2} + {b^2}} \right){x_1}{y_1} = 0} \\ \end{gathered}$

This is the equation of the normal to the given hyperbola at $\left( {{x_1},{y_1}} \right)$.