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).