# Equations of Tangent and Normal to the 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 that the standard equation of a parabola with vertex at origin $\left( {0,0} \right)$ can be written as
${y^2} = 4ax\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Since the point $\left( {{x_1},{y_1}} \right)$ lies on the given parabola, it must satisfy equation (i). So we have
${y_1}^2 = 4a{x_1}\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Now differentiating equation (i) on both sides with respect to $x$, we have
$2y\frac{{dy}}{{dx}} = 4a \Rightarrow \frac{{dy}}{{dx}} = \frac{{2a}}{y}$

If $m$ represents the slope of the tangent at the given point $\left( {{x_1},{y_1}} \right)$, then
$m = {\frac{{dy}}{{dx}}_{\left( {{x_1},{y_1}} \right)}} = \frac{{2a}}{{{y_1}}}$

The equation of the tangent at the given point $\left( {{x_1},{y_1}} \right)$ is
$\begin{gathered} y – {y_1} = \frac{{2a}}{{{y_1}}}\left( {x – {x_1}} \right) \\ \Rightarrow {y_1}y – {y_1}^2 = 2ax – 2a{x_1} \\ \Rightarrow {y_1}y – 4a{x_1} = 2ax – 2a{x_1} \\ \Rightarrow {y_1}y = 2ax + 2a{x_1} \\ \Rightarrow \boxed{{y_1}y = 2a\left( {x + {x_1}} \right)} \\ \end{gathered}$

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

The slope of normal at $\left( {{x_1},{y_1}} \right)$ is $– \frac{1}{m} = – \frac{{{y_1}}}{{2a}}$

The equation of normal at the point $\left( {{x_1},{y_1}} \right)$ is $y – {y_1} = – \frac{{{y_1}}}{{2a}}\left( {x – {x_1}} \right)$
$\begin{gathered} \Rightarrow 2ay – 2a{y_1} = – {y_1}x + {x_1}{y_1} \\ \Rightarrow \boxed{{y_1}x + 2ay – 2a{y_1} – {x_1}{y_1} = 0} \\ \end{gathered}$

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