A helicopter is flying along the curve giv en by$\mathrm{y}={\mathrm{x}}^{3/2}=7, (\mathrm{x}\ge 0).$ A solider positioned at the point $\left(\frac{1}{2},7\right)$  wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is

$$\begin{gathered} y-x^{\frac{3}{2}}=7\to \left(1\right) \\ \text{ The nearest distance}\text{ is }AB \end{gathered}$$
$$\begin{gathered} \text{ Then slope of line }L\text{ is }{m}_1=\frac{3}{2}\sqrt{x} \\ \text{ and slope of line }AB{m}_2=\frac{7-y}{\frac{1}{2}-x}=\frac{2(7-y)}{1-2x} \\ \text{ As }AB\text{ and line }L\text{ is perpendicular } \\ \begin{array}{l}\left(\frac{3}{2}\sqrt{x}\right)\left(\frac{2(7-y)}{1-2x}\right)=-1\\ \Rightarrow 3\sqrt{x}\cdot x^{\frac{3}{2}}=-2x+1\\ \Rightarrow 3x^2+2x-1=0 \\ \Rightarrow x=\frac{1}{3},\text{ and }x=-1\text{ which is rejected }\end{array} \end{gathered}$$

$$\begin{gathered} \text{ so, curve intersect at }y={\left(\frac{1}{3}\right)}^{3/2}+7 \\ \begin{array}{l}\left|AB\right|=\sqrt{{\left(\frac{1}{2}-\frac{1}{3}\right)}^2+{\left(7-7-{\left(\frac{1}{3}\right)}^{3/2}\right)}^2}\\ \Rightarrow \left|AB\right|=\sqrt{{\left(\frac{1}{6}\right)}^2+{\left(\frac{1}{3}\right)}^3}=\sqrt{\frac{7}{108}}\\ =\frac{1}{6}\sqrt{\frac{7}{3}}\end{array} \end{gathered}$$