JEE Main 2019 — Application of Derivatives Question with Solution

We have

 $y=7+x^{\frac{3}{2}}$

$\Rightarrow \frac{dy}{dx}=\frac{3}{2}\sqrt{x}$

Let the point on curve be $P\left(x_1, 7+x_1^{\frac{3}{2}}\right)$ and given point be $A\left(\frac{1}{2},7\right)$.

For nearest point normal at $P$ passes through $A$.

So, the slope of line $AP=$ Slope of normal at $P$

$\Rightarrow \frac{x_1^{\frac{3}{2}}}{x_1-\frac{1}{2}}={{\left.-\frac{dx}{dy}\right|}_{\left(x_1,y_1\right)}}_{ }=-\frac{2}{3\sqrt{x_1}}$

$\Rightarrow 3x_1^2=1-2x_1$

$\Rightarrow 3x_1^2+2x_1-1=0$

$\Rightarrow \left(x_1+1\right)\left(3x_1-1\right)=0$

$\Rightarrow x_1=\frac{1}{3} (x_1=-1$, is not possible as $x_1>0)$

Hence, point $P$ is $\left(\frac{1}{3},7+\frac{1}{3\sqrt{3}}\right)$.

So, $AP=\sqrt{\frac{1}{36}+\frac{1}{27}}=\frac{1}{6}\sqrt{\frac{7}{3}}$ units.