The given equation of the curve is $y^{2}=4ax ...(1)$
Differentiating both sides of $(1)$ with respect to $x$, we get
$2 y \frac{d y}{d x}=4 a ;$
$\Rightarrow \frac{d y}{d x}=\frac{4 a}{2 y}=\frac{2 a}{y} ...(2)$
If $\psi$ be the angle which the tangent to the curve at $( x, y )$
makes with the positive direction of $x$-axis then tan
$\psi=\frac{d y}{d x}$ or $\tan \psi=$
$\frac{2 a}{y} ...(3),$[using (2)]
At $x =a,$ then from (1),
$y^{2}=4 a \cdot a=4 a^{2}$
$\Rightarrow y=\pm 2 a$
Hence, we get two points $(a, 2 a)$ and $(a,-2 a)$ on the
curve.
At $(a, 2 a) x=a, y=2a$ and let $\psi=\psi_{1}$
$\therefore $ from $(3),\tan \psi_{1}=\frac{2 a}{2 a}=1=\tan 45^{\circ};$ $\Rightarrow \psi_{1}=45^{\circ}$
At $(a,-2 a), x=a, y=-2 a$ 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 $(1)$ at $(a, 2 a)$ and $(a,-2 a)=\psi_{2}-\psi_{1}=135^{\circ}-45^{\circ}$ $=90^{\circ}$.
This shows that the tangent lines to $(1)$ at $(a, 2 a)$ and $(a,-2 a)$ are perpendicular to each other.