Question

The closest point on the parabola \[ y=x^2+7x+2 \] to the straight line \[ y=3x-2 \] is

• A. \((-1,-4)\)
• B. \((1,10)\)
• C. \((-2,-8)\)
• D. \((0,2)\)

Hint:

To find the point on a curve nearest to a line, substitute the curve equation into the line-distance formula and minimize the resulting expression. Often it simplifies into a perfect square.

Answer 1

The Correct Option is C

Concept: The point on the parabola closest to a line minimizes the perpendicular distance. Express distance squared, use calculus or complete the square to find the minimum.

Step 1: Parabola: \(y=x^2+7x+2\). Line: \(3x-y-2=0\). Distance \(d = |3x - y - 2|/\sqrt{10}\). Substitute y: \(3x - (x^2+7x+2) - 2 = -x^2 -4x -4 = -(x+2)^2\). So \(d = (x+2)^2/\sqrt{10}\).

Step 2: Minimum when \((x+2)^2=0 \Rightarrow x=-2\). Then \(y = 4-14+2 = -8\). Point: \((-2,-8)\).

Step 3: Write the final answer. \(\boxed{(-2,-8)}\)