the point on the curve x^2+2y=0 that is nearest the point (0,-1/2) occurs where y is....

Question

A: 1/2

B: 0

C: -1/2

D: -1

E: None of the above

Answer 1

Hi Ricardo M.,

You can "game the system" easily, without even calculating, when you are asked a multiple choice question like this. You would first rearrange to y=-(1/2)x^2, note that this is a downwards-pointing parabola, and reason that if there is a single point.(i.e. a single choice), it must be symmetric with respect to the y-axis. Only (0,0) satisfies this as a single unique point.

If you wanted to be more rigorous, you could reason that a closest point would be connected by a segment perpendicular to the intersection with the curve, and solve; or, you could set up an expression for the distance from the point to the curve, and minimize it (as a derivative with respect to dx).

Yet another way, would be to reason that the given point is the focus of the parabola, hence closest to the vertex.

-- Cheers, -- Mr. d.

Answer 2

I suggest minimizing the distance that you get from applying the distance formula.

x² + 2y = 0 ⇒

y = -x²/2

So you want to minimize the distance between (x, -x²/2) and (0, -1/2). The square of the distance between the points is

(x - 0)² + (-x²/2 + 1/2)² = x² + (1/2 - x²/2)²

It should be sufficient to minimize the square of the distance, so take its derivative, set it equal to zero, solve for x, and then y = -x²/2. That should work.