Question

Consider the parabola $y=x^2+7x+2$ and the straight line $y=3x-3$.Find the coordinates of the point on the parabola which is closest to the straight line.

• A. (0, 2)
• B. (-2, -8)
• C. (-7, 2)
• D. (1, 10)

Answer 1

The correct option is B (-2, -8)

Let $(h,k)$ be the point closest to the line.

So, the tangent at $(h,k)$ must be parallel to the given line

$y=x^2+7x+2$

$\therefore \frac{dy}{dx}=2x+7$

Slope of line $y=3x-3$ is $3$

$\therefore 2h+7=3$

$\therefore h=-2$

$\therefore k=h^2+7h+2=4-14+2=-8$

So, the answer is option (B).