Find the Equation of the Tangent Line to the Hyperbola

Show that the \left( {a\sec \theta ,b\tan \theta } \right) always lies on the hyperbola \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1. Find the equation of the tangent and normal to the hyperbola \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 at the point \left( {a\sec \theta ,b\tan \theta } \right).

We have standard equation of a hyperbola:

\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Putting x = a\sec \theta and y = b\tan \theta in equation (i), we have

\begin{gathered} \frac{{{{\left( {a\sec \theta } \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {b\tan \theta } \right)}^2}}}{{{b^2}}} = 1 \\ \Rightarrow \frac{{{a^2}{{\sec }^2}\theta }}{{{a^2}}} - \frac{{{b^2}{{\tan }^2}\theta }}{{{b^2}}} = 1 \\ \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\ \end{gathered}

Which is true for all values of \theta , so the point \left( {a\sec \theta ,b\tan \theta } \right) always lies on the hyperbola (i).

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{{dy}}{{dx}} = \frac{{{b^2}x}}{{{a^2}y}} \\ \end{gathered}

Let m be the slope of the tangent at the given point \left( {a\sec \theta ,b\tan \theta } \right), then

m = {\frac{{dy}}{{dx}}_{\left( {a\cos \theta ,b\sin \theta } \right)}} = \frac{{{b^2}\left( {a\sec \theta } \right)}}{{{a^2}\left( {b\tan \theta } \right)}} = \frac{{b\sec \theta }}{{a\tan \theta }}

The equation of the tangent at the given point \left( {a\sec \theta ,b\tan \theta } \right) is

\begin{gathered} y - b\tan \theta = \frac{{b\sec \theta }}{{a\tan \theta }}\left( {x - a\sec \theta } \right) \\ \Rightarrow \frac{{\tan \theta }}{b}\left( {y - b\tan \theta } \right) = \frac{{\sec \theta }}{a}\left( {x - a\sec \theta } \right) \\ \Rightarrow \frac{{y\tan \theta }}{b} - {\tan ^2}\theta = - \frac{{x\sec \theta }}{a} - {\sec ^2}\theta \\ \Rightarrow \frac{x}{a}\sec \theta - \frac{y}{b}\tan \theta = {\sec ^2}\theta - {\tan ^2}\theta \\ \Rightarrow \frac{x}{a}\sec \theta - \frac{y}{b}\tan \theta = 1 \\ \end{gathered}

This is the equation of the tangent to the given hyperbola at \left( {a\sec \theta ,b\tan \theta } \right).

The slope of the normal at \left( {a\sec \theta ,b\tan \theta } \right) is  - \frac{1}{m} = - \frac{{a\tan \theta }}{{b\sec \theta }}

The equation of the normal at the point \left( {a\sec \theta ,b\tan \theta } \right) is

\begin{gathered} y - b\tan \theta = - \frac{{a\tan \theta }}{{b\sec \theta }}\left( {x - a\sec \theta } \right) \\ \Rightarrow \frac{b}{{\tan \theta }}\left( {y - b\tan \theta } \right) = \frac{a}{{\sec \theta }}\left( {x - a\sec \theta } \right) \\ \Rightarrow \frac{a}{{\sec \theta }}x - \frac{b}{{\tan \theta }}y = {a^2} - {b^2} \\ \end{gathered}

This is the equation of the normal to the given hyperbola at \left( {a\sec \theta ,b\tan \theta } \right).