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).