Question

Find the co-ordinates of a point on the parabola $y=x^2+7x+2$ which is closest to the straight line $y=3x-3$.

• A. $(1,10)$
• B. $(-1,-4)$
• C. $(2,20)$
• D. $(-2,-8)$

Answer 1

The correct option is D $(-2,-8)$

Let the point $P(x,y)$ be on the parablola $y=x^2+7x+2$

Given line is

$y=3x-3or3x-y-3=0$

Distance of P from the line

$D=\left|\frac{3x-y-3}{\sqrt{\left(10\right)}}\right|=\left|\frac{3x-(x^2+7x+2)-3}{\sqrt{\left(10\right)}}\right|=\left|\frac{-x^2-4x-5}{\sqrt{\left(10\right)}}\right|$

$D=\left|\frac{x^2+4x+5}{\sqrt{\left(10\right)}}\right|=\left|\frac{{(x+2)}^2+1}{\sqrt{\left(10\right)}}\right|$

$=\frac{{(x+2)}^2+1}{\sqrt{\left(10\right)}}$ as $\frac{N^r}{D^r}$ is+ive

$\frac{dD}{dx}=\frac{2(x+2)}{\sqrt{\left(10\right)}}$

For maxima or minima,

$\frac{dD}{dx}=0$

$\Rightarrow x=-2$

$\frac{d^{2}D}{d{x}^2}=\frac{2}{\sqrt{\left(10\right)}}=+ive$

$\Rightarrow y=-8$

Hence,the required point is $(-2,-8)$.