If the tangent to the curve

Given that the tangent to the curve $x=a t^{2}, y=2 a t$ is perpendicular to $x$-axis.

Differentiating both w.r.t. $t$,

$\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{at}, \frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{a}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{2 \mathrm{a}}{2 \mathrm{at}}=\frac{1}{\mathrm{t}}$

From $y=2 a t, t=\frac{y}{2 a}$

$\Rightarrow$ Slope of the curve $=\frac{2 a}{y}$

$\Rightarrow \frac{2 a}{y}=0$

$\Rightarrow a=0$

Then point of contact is $(0,0)$.