A helicopter is flying along the curve given by

Question:

A helicopter is flying along the curve given by $y-x^{3 / 2}=7,(x \geq 0)$. 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:

 

  1. (1) $\frac{\sqrt{5}}{6}$

  2. (2) $\frac{1}{3} \sqrt{\frac{7}{3}}$

  3. (3) $\frac{1}{6} \sqrt{\frac{7}{3}}$

  4. (4) $\frac{1}{2}$


Correct Option: , 3

Solution:

$f(x)=y=x^{3 / 2}+7$

$\Rightarrow \quad \frac{d y}{d x} \Rightarrow \frac{3}{2} \sqrt{x}>0$

$\Rightarrow f(x)$ is increasing function $\forall x>0$

Let $P\left(x_{1}, x_{1}^{3 / 2}+7\right)$

$m_{\mathrm{TP}}=m_{\mathrm{at} \mathrm{P}}=-1$

$\Rightarrow\left(\frac{x_{1}^{3 / 2}}{x_{1}-\frac{1}{2}}\right) \times \frac{3}{2} x_{1}^{\frac{1}{2}}=-1$

$\Rightarrow \quad-\frac{2}{3}=\frac{x_{1}^{2}}{x_{1}-\frac{1}{2}}$

$\Rightarrow \quad-3 x_{1}^{2}=2 x_{1}-1 \Rightarrow 3 x_{1}^{2}+2 x_{1}-1=0$

$\Rightarrow \quad 3 x_{1}^{2}+3 x_{1}-x_{1}-1=0$

$\Rightarrow \quad 3 x_{1}\left(x_{1}+1\right)-1\left(x_{1}+1\right)=0$

$\Rightarrow \quad x_{1}=\frac{1}{3}$ $\left(\because x_{1}>0\right) \Rightarrow P\left(\frac{1}{3}, 7+\frac{1}{3 \sqrt{3}}\right)$

$\mathrm{TP}=\sqrt{\frac{1}{27}+\frac{1}{36}}=\frac{1}{6} \sqrt{\frac{7}{3}}$

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