The tangent to ${\mathrm{y}}^2=\mathrm{ax}$ makes an angle of 45° with x–axis. Then its point of contact is

${\mathrm{y}}^2=\mathrm{ax}$ is comparing with

$$\begin{gathered} {\mathrm{y}}^2=4\mathrm{Ax} \\ \Rightarrow \mathrm{A}=\mathrm{a}/4 \\ Given \mathrm{\theta }={45}^0 \\ Now slope of \tan gent \mathrm{m}=\tan {45}^0=1 \end{gathered}$$

Now point of contact  is $=\left(\frac{\mathrm{A}}{{\mathrm{m}}^2},\frac{2\mathrm{A}}{\mathrm{m}}\right)=\left(\frac{a}{4} ,\frac{a}{2}\right)$