A tangent to the curve $x=a t^{2}, y=2 a t$ is perpendicular to $X$ axis, then the point of contact is
A tangent to the curve $x=a t^{2}, y=2 a t$ is perpendicular to $X$ axis, then the point
of contact is
$(0,-a)$
$(0,0)$
$(0,2 a)$
$(0, a)$
Solution
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 $\mathrm{y}=0$