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 $$\left( {{x_1},{y_1}} \right)$$.