If the normal at 'P' on ${\mathrm{y}}^2=4\mathrm{ax}$ cuts the axis of the parabola in G and S is the focus then SG =

$\mathrm{Given} \mathrm{parabola} {\mathrm{y}}^2=4\mathrm{ax}$

Let $\mathrm{p}({\mathrm{x}}_1, {\mathrm{y}}_1)$ is a point on the parabola 

diff w.r.to x on both sides 

$$\begin{gathered} 2\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=4\mathrm{a} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{a}}{\mathrm{y}} \end{gathered}$$

slope of normal is $\raisebox{1ex}{$-{\mathrm{y}}_1$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{a}$}\right.$ 

Equation of normal is $\mathrm{y}-{\mathrm{y}}_1=\frac{-{\mathrm{y}}_1}{2\mathrm{a}}\left(\mathrm{x}-{\mathrm{x}}_1\right)$

 it meets x-axis at G Then $\mathrm{y}=0$

$$\begin{gathered} -\cancel{{\mathrm{y}}_1}=\frac{\cancel{{\mathrm{y}}_1}}{2\mathrm{a}}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \mathrm{x}={\mathrm{x}}_1+2\mathrm{a} \\ \therefore \mathrm{G}=\left({\mathrm{x}}_1+2\mathrm{a}, 0\right) \\ and \mathrm{S}=focus =(\mathrm{a}, 0) \\ Now \mathrm{SG}=\sqrt{{\left({\mathrm{x}}_1+2\mathrm{a}-\mathrm{a}\right)}^2+0} \\ \Rightarrow \mathrm{SG}=\left|{\mathrm{x}}_1+\mathrm{a}\right|=\mathrm{SP} \end{gathered}$$