Find the coordinates of a point on the parabola $y=x^{2}+7 x+2$ which is closest to the strainght line $y=3 x-3$.
Let coordinates of the point on the parabola be $(x, y)$. Then,
$y=x^{2}+7 x+2$ ......(1)
Let the distance of a point $\left(x,\left(x^{2}+7 x+2\right)\right)$ from the line $y=3 x-3$ be $S .$ Then,
$S=\left|\frac{-3 x+\left(x^{2}+7 x+2\right)+3}{\sqrt{10}}\right|$
$\Rightarrow \frac{d S}{d t}=\frac{-3+2 x+7}{\sqrt{10}}$
For maximum or minimum values of $S$, we must have
$\frac{d S}{d t}=0$
$\Rightarrow \frac{-3+2 x+7}{\sqrt{10}}=0$
$\Rightarrow 2 x=-4$
$\Rightarrow x=-2$
Now,
$\frac{d^{2} S}{d t^{2}}=\frac{2}{\sqrt{10}}>0$
So, the nearest point is $\left(x,\left(x^{2}+7 x+2\right)\right)$.
$\Rightarrow(-2,4-14+2)$
$\Rightarrow(-2,-8)$
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