A tangent to the curve \(x=a t^{2}, y=2 a t\) is perpendicular to \(X\) axis, then the point
of contact is

Given equation of curve represents parabola \(y^{2}=4 a x\)
Given is that tangent is \(\perp\) er to \(X\)-axis. Therefore point of contact is vertex of parabola which is origin.
This problem can also be solved as
We have \(x=a t^{2}\) and \(y=2\) at
\(\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{at} \quad\) and \(\quad \frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{a}\)
\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{2 \mathrm{at}}=\frac{1}{\mathrm{t}} \Rightarrow\) Slope of tangent \(=\frac{1}{\mathrm{t}}\). Since tangent is Perpendicular to \(\mathrm{X}\) axis, it is parallel to \(\mathrm{Y}\) axis i.e. it's slope is indefinite. It means \(\mathrm{t}=0 \Rightarrow \mathrm{x}=0\) and \(\mathrm{y}=0\)