Question

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

• A. $(-2,-8)$
• B. $(2,8)$
• C. $(0,2)$
• D. $(0,-2)$

Answer 1

Answer: (A)

$(-2,-8)$

Given equation of parabola is $y=x^2+7x+2$
Let the point $P(x_1,y_1)$ be on the parabola
$y_1=x_1^2+7x_1+2$

Distance of the line $3x-y-3=0$ from $P(x_1,y_1)$ is
$D=\left|\frac{3x_1-y_1-3}{\sqrt{(}10)}\right|$
$=\left|\frac{3x_1-(x_1^2+7x_1+2)-3}{\sqrt{(}10)}\right|$

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

$\frac{dD}{dx}=\frac{2(x+2)}{\sqrt{(}10)}=0,$ $\therefore x=-2$
$\frac{d^{2}D}{d{x}^2}=\frac{2}{\sqrt{(}10)}>0$
Hence, $D$ has a minimum at $x=-2$
$\Rightarrow y=-8$

So, the point is $(-2,-8)$.

Hence, option A.