JEE Main 2019 — Application of Derivatives Question with Solution

Let $P$ be the nearest to $\left(\frac{3}{2}, 0\right),$ then normal at $P$ will pass through $\left(\frac{3}{2}, 0\right).$

Any point on the parabola $y^2=4ax$ can be taken as $\left(a{t}^2, 2at\right)$

Let, co-ordinates of $P$ on the curve $y=\sqrt{x}, \Rightarrow y^2=x$ can be taken as $\left(\frac{t^2}{4}, \frac{t}{2}\right)$


The equation of the normal to the parabola $y^2=4ax$ at a point $\left(a{t}^2, 2at\right)$ is $y+tx=2at+a{t}^3.$

Hence, the equation of normal to $y=\sqrt{x}$ or $y^2=x$ at $\left(\frac{t^2}{4}, \frac{t}{2}\right)$ is $y+tx=\frac{t}{2}+\frac{t^3}{4}$

This line passes through $\left(\frac{3}{2}, 0\right),$ hence 
$\frac{3t}{2}=\frac{t}{2}+\frac{t^3}{4}$

$\Rightarrow t^3-4t=0$

$\Rightarrow t\left(t^2-4\right)=0$

$\Rightarrow t=0, 2, -2$ but $-2$ is rejected as we have to take point in the first quadrant.

Thus, the points are $\left(0, 0\right)$ or $\left(1, 1\right)$ and the distances of these points from $\left(\frac{3}{2}, 0\right)$ are respectively

$\sqrt{{\left(\frac{3}{2}-0\right)}^2+{\left(0-0\right)}^2}=\frac{3}{2}$ and $\sqrt{{\left(\frac{3}{2}-1\right)}^2+{\left(1-0\right)}^2}=\frac{\sqrt{5}}{2}$ units.

Hence, nearest point is $\left(1,1\right)$ and the minimum distance is $\frac{\sqrt{5}}{2}$ units.