Question

An Aapache helicopter of 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.

Answer 1

Given that an apache helicopter of enemy is flying along the curve given by $=x^2+7$

A soldier placed at $(3,7)$ wants to shoot down the helicopter when it is nearest to him.

Now, the distance of the point from the soldier is $\sqrt{(x-3{)}^2+(y-7{)}^2}$

Since the point lies on the curve $y=x^2+7$..............1
$S=\sqrt{(x-3{)}^2+(x^2+7-7{)}^2}$

$\to S=\sqrt{x^4+x^2-6x+9}$
When the distance is maximum/minimum,
$\frac{dS}{dx}=0$
$\to \frac{(4x^3+2x-6)}{\sqrt{(x^4+x^2-6x+9)}}=0$
$\to 4x^3+2x-6=0$
$\to 2x^3+x-3=0$
$\to (x-1)\times (2x^2+2x+3)$
The solutions to the above equation are
$x=1,-0.5\pm i\times 0.5\sqrt{5}$
Since the solution cannot have any complex roots.

Hence, $x=1$ is the abscissa of the nearest point to the soldier.

From equation $1$, we get
$y=1^2+7$

$\to y=8$
So, the nearest point is $(1,8)$
Now, nearest distance $S=\sqrt{(1-3{)}^2+(8-7{)}^2}$

$\to S=\sqrt{(-2{)}^2+(1{)}^2}$

$\to S=\sqrt{4+1}$

$\to S=\sqrt{5}$ units