Find the coordinates of a point on the parabola

Solution:

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 $\left(x,\left(x^{2}+7 x+2\right)\right)$.

$\Rightarrow(-2,4-14+2)$

$\Rightarrow(-2,-8)$