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An enemy fighter jet is flying along the curve, given by $y=x^2+2$. A soldier is placed at $(3,2)$ wants to shoot down the jet when it is nearest to him. Then, the nearest distance is
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The correct answer is:
$\sqrt{5}$ units
Let $P(x, y)$ be the position of the jet and the
solider is placed at $A(3,2)$.
$\Rightarrow \quad A P=\sqrt{(x-3)^2+(y-2)^2}$
As, $\quad y=\left(x^2+2\right) \Rightarrow x^2=y-2$
$\Rightarrow(A P)^2=(x-3)^2+x^4=z$
(say)
$\Rightarrow \quad \frac{d z}{d x}=2(x-3)+4 x^3$ and $\frac{d^2 z}{d x^2}=2+12 x^2$
For local points of maxima and minima
$\begin{aligned} & \frac{d z}{d x}=0 \Rightarrow 2(x-3)+4 x^3=0 \\ & x=1 \text { and } \frac{d^2 z}{d x^2}(\text { at } x=1)=14>0\end{aligned}$
$\therefore z$ is minimum, whe $x=1, y=1+2=3$
Also, minimum distance
$=\sqrt{(1-3)^2+(3-2)^2}=\sqrt{5}$ units
solider is placed at $A(3,2)$.
$\Rightarrow \quad A P=\sqrt{(x-3)^2+(y-2)^2}$
As, $\quad y=\left(x^2+2\right) \Rightarrow x^2=y-2$
$\Rightarrow(A P)^2=(x-3)^2+x^4=z$
(say)
$\Rightarrow \quad \frac{d z}{d x}=2(x-3)+4 x^3$ and $\frac{d^2 z}{d x^2}=2+12 x^2$
For local points of maxima and minima
$\begin{aligned} & \frac{d z}{d x}=0 \Rightarrow 2(x-3)+4 x^3=0 \\ & x=1 \text { and } \frac{d^2 z}{d x^2}(\text { at } x=1)=14>0\end{aligned}$
$\therefore z$ is minimum, whe $x=1, y=1+2=3$
Also, minimum distance
$=\sqrt{(1-3)^2+(3-2)^2}=\sqrt{5}$ units
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