# A helicopter of the enemy is flying along the curve given by  $y=x^2+7$. A soldier, placed at (3, 7), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.

Curve  $y=x^2+7$ and  $P(x,x^2+7)$ is cordinate and  $Q(3,7)$.

Distance  $PQ=\sqrt{(x-3)^2+(x^2-7+7)^2}$

$(PQ)^2=(x-3)^2+x^4$

Differentiate wrt x,

$f(x)=\frac{\mathrm{d} }{\mathrm{d} x}(PQ)^2=2(x-3)+4x^3=0$

$2x-6+4x^3=0$

$x-3+2x^3=0$

$2(x-1)(2x^2+2x+3)=0$

In the quadratic part, D<0. Therefore no roots.

$\therefore x=1$  is the root of the equation.

$f'(1)<0$

Minima is at x = 1

$PQ=\sqrt{(x-3)^2+(x^2)^2}$

At x=1,

$PQ=\sqrt{4+1}=\sqrt{5}$

Therefore, the minimum distance is  $\sqrt{5}$.

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