Question

Anyone know?
Find a point on the curve defined by function y = x2 that has the minimum distance to the
line defined by y = 2x - 4.
(a) Formulate the problem as a non-linear optimization problem.

Answer 1

Let (p, q) be any point on the curve y = x^2.
Then q = p^2. So the point on the curve is (p, p^2).

Let (r, s) be any point on the line y = 2x-4
Then s = 2r-4. So the point on the line is (r, 2r-4).

The square of the distance between these two points is:
Dsquare = (p-r)^2 + (p^2 - 2r + 4)^2 ----- (1)
We need to minimize Dsquare (minimizing the square of the distance is equivalent to minimizing the distance).
But there are two variables, p and r. We need to find a relationship between p and r so we can eliminate one of them.

To measure the distance from a point to a line we drop a perpendicular from the point to the line and measure the distance. Therefore, the line joining the point on the straight line and the point on the curve must be perpendicular to each other. The slope of the line y = 2x - 4 is 2. Therefore, the slope of the perpendicular line is -1/2.
What is the slope of the line joining (p,p^2) and (r, 2r-4) ?
Equate that to -1/2, find r in terms of p, sub in (1) and minimize Dsquare.

Answer 2

What is the slope of the line joining (p,p^2) and (r, 2r-4) ?
slope = (p^2-2r+4) / (p-r) = -1/2
cross multiply:
2p^2 - 4r + 8 = r - p
2p^2 + p + 8 = 5r
r = 1/5 * (2p^2 + p + 8).
Substitute the above in equation (1) and minimize Dsquare.