Question

A helicopter is flying along the curve given by $y-x^{\frac{3}{2}}=7,(x\ge 0)$. A soldier positioned at the point $(\frac{1}{2},7)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :

• A. $\frac{\sqrt{5}}{6}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}\sqrt{\frac{7}{3}}$
• D. $\frac{1}{6}\sqrt{\frac{7}{3}}$

Answer 1

The correct option is D $\frac{1}{6}\sqrt{\frac{7}{3}}$

$y-x^{\frac{3}{2}}=7\cdots \left(1\right)$


The nearest distance will be $AB$ as shown in figure
Then slope of line $L$ is $m_1=\frac{3}{2}\sqrt{x}$
and slope of line $AB$ $m_2=\frac{7-y}{\frac{1}{2}-x}=\frac{2(7-y)}{1-2x}$
As $AB$ and line $L$ is perpendicular
$\left(\frac{3}{2}\sqrt{x}\right)\left(\frac{2(7-y)}{1-2x}\right)=-1$
$\Rightarrow 3\sqrt{x}\cdot x^{\frac{3}{2}}=-2x+1$
$\Rightarrow 3x^2+2x-1=0$
$\Rightarrow x=\frac{1}{3},$ and $x=-1\text{which is rejected}$
so, curve intersect at $y={\left(\frac{1}{3}\right)}^{3/2}+7$
Now,
$|AB|=\sqrt{{(\frac{1}{2}-\frac{1}{3})}^2+{(7-7-{\left(\frac{1}{3}\right)}^{3/2})}^2}$
$\Rightarrow |AB|=\sqrt{{\left(\frac{1}{6}\right)}^2+{\left(\frac{1}{3}\right)}^3}=\sqrt{\frac{7}{108}}$
$=\frac{1}{6}\sqrt{\frac{7}{3}}$