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

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$\frac{1}{6} \sqrt{\frac{7}{3}}$

$\frac{\sqrt{5}}{6}$

$\frac{1}{3} \sqrt{\frac{7}{3}}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} y = 7 + x^{3 / 2} \\ \frac{dy}{dx} = \frac{3}{2} x^{1 / 2} > 0 \end{array}$
is increasing function
Let $p (t, 7 + t^{3 / 2})$ be point on curve and as $Q (\frac{1}{2}, 7)$
$\begin{aligned} PQ = \sqrt{(t - 1 / 2)^{2} + {(t^{3 / 2})}^{2}} \\ \frac{d}{dt} (PQ)^{2} = 2 (t - 1 / 2) + 3 t^{2} \end{aligned}$
It has minimum at $t = \frac{1}{3}$
$PQ = \sqrt{\frac{1}{36} + \frac{1}{27}}$