• # question_answer A helicopter is flying along the curve given by $y-{{x}^{3/2}}=7,(x\ge 0).$ A soldier 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: A) $\frac{\sqrt{5}}{6}$ B) $\frac{1}{3}\sqrt{\frac{7}{3}}$ C) $\frac{1}{6}\sqrt{\frac{7}{3}}$ D) $\frac{1}{2}$

Solution :

 $y={{x}^{3/2}}-2$     $\frac{dy}{dx}=\frac{3}{2}\sqrt{x}$ Slope of normal $=-\frac{2}{3\sqrt{x}}$ Let point is$({{x}_{1}},x_{1}^{3/2}-2)$ $\therefore$Normal$y-(x_{1}^{3/2}-2)=\frac{-2}{3\sqrt{{{x}_{1}}}}(x-{{x}_{1}})$
 Now put$(1,7)$and solve it. $\Rightarrow$           ${{x}_{1}}=\frac{1}{3}$ $\therefore$      $P\Rightarrow \left( \frac{1}{3},7+\frac{1}{3\sqrt{3}} \right),A\Rightarrow (1,7)$ $\therefore$      $AD=\frac{1}{6}\sqrt{\frac{7}{3}}.$

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