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

{y^2}  = 4ax\,\,\,{\text{ -  -  - }}\left( {\text{i}} \right)


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

{y_1}^2  = 4a{x_1}\,\,\,{\text{ -  -  - }}\left( {{\text{ii}}} \right)


Now differentiating equation (i) 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 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}}}


Equation of 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 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)

\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).

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