Ira's problem

Sanderson Smith (
Tue, 25 May 1999 19:30:06 -0700

Hi group....

I'll take a shot at presenting an approach on how to solve Ira's problem.

I'm assuming that the problem was to find the minimum distance from the parabola y = x^2+4 to the line y = x-3.

Let P1= (w, w^2+4) be a point on the parabola. Now, the distance from P1 to the line y = x-3 would be the perpendicular
distance, so the distance would be along a line with slope =-1.

The line L through P1 with slope -1 has equation y-(w^2+4) = -1(x-w).

Let P2 = (z,z-3) be the point on at which L intersects the line y = x-3.

Then, (z-3)-(w^2+4) = -1(z-w)

Solving this for z yields z = (1/2)(w^2+w+7) = .5w^2 + .5w + 3.5.

So the point P2 = (.5w^2+.5w+3.5, .5w^2 + .5w +.5)

We want to minimize the distance from P1 to P2. Since both points now have coordinates expressed only in terms of the variable
w, use the distance formula to express the distance D from P1 to P2, calculate the derivative dD/dw, set it equal to zero, and
solve for w. This should produce the value of w that minimizes the distance, D.

Perhaps there is an easier way, but this is what came to mind.

Warm regards to all.
-Keep smiling.