The given equation of the curve is $y^{2}=4 a x$
Differentiating both sides of $(1)$ with respect to $\mathrm{x}$, we get
$$
2 y \frac{d y}{d x}=4 a ; \Rightarrow \frac{d y}{d x}=\frac{4 a}{2 y}=\frac{2 a}{y}
$$
If $\psi$ be the angle which the tangent to the curve at ( $\mathrm{x}, \mathrm{y})$ makes with the positive direction of $x$-axis then $\tan \psi=\frac{d y}{d x}$ or
$$
\tan \psi=\frac{2 \mathrm{a}}{\mathrm{y}} \quad \ldots .(3), \quad[\mathrm{using}(2)]
$$
At $x=a$, then from $(1), y^{2}=4 a \cdot a=4 a^{2} \Rightarrow y$ $=\pm 2 \mathrm{a}$

Hence, we get two points $(\mathrm{a}, 2 \mathrm{a})$ and $(\mathrm{a},-$ 2a) on the curve.
$\mathrm{At}(\mathrm{a}, 2 \mathrm{a}) \mathrm{x}=\mathrm{a}, \mathrm{y}=2 \mathrm{a}$ and let $\psi=\psi_{1}$
$$
\begin{array}{r}
\therefore \text { from (3) } \tan \psi_{1}=\frac{2 a}{2 a}=1=\tan 45^{\circ} \\
\Rightarrow \psi_{1}=45^{\circ} .
\end{array}
$$
$\mathrm{At}(\mathrm{a},-2 \mathrm{a}), \mathrm{x}=\mathrm{a}, \mathrm{y}=-2 \mathrm{a}$ and let $\psi=\psi_{2}$
$\therefore$ from (3), $\tan \psi_{2}=\frac{2 a}{-2 a}=-1=\tan 135^{\circ}$;
or $\psi_{2}=135^{\circ}$.
Hence the required angle between tangents to $(1) \operatorname{at}(\mathrm{a}, 2 \mathrm{a})$ and $(\mathrm{a},-2 \mathrm{a})=\psi_{2}-\psi_{1}=135^{\circ}-$ $45^{\circ}=90^{\circ} .$
This shows that the tangent lines to (1) at $(a, 2 a)$ and $(a,-2 a)$ are perpendicular to each other.