# 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 $\left( {a\sec \theta ,b\tan \theta } \right)$.