Question

The coordinates of the point on the parabola $y=x^2+7x+2$, which is nearest to the straight line $y=3x-3$ are

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

Answer 1

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

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

$y=x^2+7x+2={(x+\frac{7}{2})}^2-\frac{41}{4}$

Let point on parabola be $(t-\frac{7}{2},t^2-\frac{41}{4})$

Equation of line is $3x-y-3=0$

List of point from line $=S=\frac{3(t-\frac{7}{2})-t^2-\frac{41}{4}-3}{\sqrt{9+1}}$

$\frac{dS}{dt}=\frac{3-2t}{\sqrt{10}}=0t=\frac{3}{2}$

Point is $(\frac{3}{2}-\frac{7}{2},\frac{9}{4}-\frac{41}{4})=(-2,-8)$