# 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 $$\left( {{x_1},{y_1}} \right)$$.