India is planning to launch a rocket with a destroyer to protect earth from a comet approaching earth. Scientists want to fire the destroyer when the rocket and the comet are closest in order to have maximum impact. For this, they found the equation of the path the rocket ( $y + x - 2 = 0$ ) will follow and the path of the comet near earth
( $y = x^{2} + 3 x + 2$ ) with respect to a reference point [(0,0)]. Find the distance between the reference point and the point where the destroyer would hit the comet.
(Assume that the rocket can be controlled to reach closest point at the same time the comet reaches )

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-1

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\sqrt{2}

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0.0

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Solution

The correct option is A 1
We are given the equation of the curves. We want to find the points where the distance between the two curves is minimum. We saw that shortest distance between two curves is along the common normal. So, we will find the common normal to find the points.

Path of rocket is $y + x - 2 = 0$ . Slope of this line is -1

Slope of normal to this line will be 1

Now the slope of the line $y = x^{2} + 3 x + 2$

$\Rightarrow f^{'} (x) = 2 x + 3$

When it is the common normal, slopes of these normal should be equal.

$2 x + 3 = 1$
$\Rightarrow x = - 1$

This is the x-coordinate of the point where the scientists are expecting to hit the comet.

$y = (- 1)^{2} + 3 (- 1) + 2 = 0$

So the point on the path of comet is (-1,0). Its distance from origin would 1.

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