# Equation of Tangent and Normal to Parabola

The equations of tangent and normal to the parabola ${y^2} = 4ax$ at the point $\left( {{x_1},{y_1}} \right)$ are ${y_1}y = 2a\left( {x + {x_1}} \right)$ and ${y_1}x + 2ay - 2a{y_1} - {x_1}{y_1} = 0$ respectively.
Consider the standard equation of parabola with vertex at origin $\left( {0,0} \right)$can be written as

Since the point $\left( {{x_1},{y_1}} \right)$ lies on the given parabola, so it must satisfy equation (i), we have

Now differentiating equation (i) both sides with respect to $x$, we have

If $m$ represents the slope of tangent at the given point $\left( {{x_1},{y_1}} \right)$, then

Equation of tangent at the given point $\left( {{x_1},{y_1}} \right)$ is

This is the equation of tangent to the given parabola at $\left( {{x_1},{y_1}} \right)$.
Slope of the normal at $\left( {{x_1},{y_1}} \right)$ is $- \frac{1}{m} = - \frac{{{y_1}}}{{2a}}$
Equation of normal at the point $\left( {{x_1},{y_1}} \right)$ is $y - {y_1} = - \frac{{{y_1}}}{{2a}}\left( {x - {x_1}} \right)$

This is the equation of normal to the given parabola at $\left( {{x_1},{y_1}} \right)$.