# 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:

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

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

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

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

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

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