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A helicopter flying the curve given by $y - x^{\frac{3}{2}} = 7, (x\geq 0)$ . A soldier positioned at the point $\left (\frac{1}{2}, 7 \right )$ wants to shoot down the helicopter when its nearest to him. Then the nearest distance is:
Option 1)
$\frac{\sqrt5}{6}$
Option 2)
$\frac{1}{6}\sqrt{\frac{7}{3}}$
Option 3)
$\frac{1}{3}\sqrt{\frac{7}{3}}$
Option 4)
$\frac{1}{2}$

Answers (1)

best_answer

Equation of Normal -

Equation of normal to the curve y = f(x) at the point P(x₁, y₁) on the curve having a slope M_N is

$(y-y_{1})=M_{N}(x-x_{1})$

$=\frac{-1}{\frac{dy}{dx}_{(x_{1},y_{1})}}(x-x_{1})$

-

Length of Tangent -

$L_{T}=\frac{y}{y'}\sqrt{1+y'^{2}}$

- wherein

Where $Where\:\:y'=\frac{dy}{dx}$

Given equation of curve

$y-x^{\frac{3}{2}}=7 \: (x\geq 0)$

from the concept

$\frac{dy}{dx}=\frac{3}{2}\sqrt x$

condition for perpendicular of two line

$=>(\frac{3}{2}\sqrt x)(\frac{7-4}{\frac{1}{2}-x})=-1$

$=>(\frac{3}{2}\sqrt x)(\frac{-x^{}\frac{3}{2}}{\frac{1}{2}-x})=-1$

$=>\frac{3}{2}x^{2}=\frac{1}{2}-x$

$=>3x^{2}+2x-1=0$

$=>(x+1)(3x-1)=0$

$x=-1\: \: is \: \: \: rejected \: \: \because x\geq 0$

So,

$x=\frac{1}{3}$

$y=7+(\frac{1}{3})^{\frac{3}{2}}$

$l_{AB}=\sqrt{(\frac{1}{2}-\frac{1}{3})^{2}}=\sqrt{\frac{1}{36}+\frac{1}{27}}=\frac{1}{6}\sqrt{\frac{7}{3}}$

Option 1)

$\frac{\sqrt5}{6}$

Option 2)
$\frac{1}{6}\sqrt{\frac{7}{3}}$
Option 3)

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

Option 4)

$\frac{1}{2}$

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