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

I have answered this question in your previous post. Here is a second method:
If you wish to make use of a standard formula for the distance between a point and a straight line, the problem can be simplified a lot.
The distance of the point (p,q) from the line Ax + By + C = 0 is given by the formula: \[
\frac{|Ap+Bq+C|}{\sqrt{A^2+B^2}}
\]The straight line here is y = 2x - 4 or 2x - y - 4 = 0.
Therefore, A = 2, B = -1, C = -4
The distance S of the point (p,q) from 2x - y - 4 = 0 is: \[
S = \frac{|2p+(-1)q+(-4)|}{\sqrt{(2)^2+(-1)^2}} =
\frac{|2p-q-4|}{\sqrt{5}}
\]The point (p,q) lies on the parabola y = x^2. Therefore, q = p^2.
\[
S = \frac{|2p-p^2-4|}{\sqrt{5}} \\
\frac{dS}{dq} = \frac{1}{\sqrt{5}}(2-2p) = 0 \\
2 = 2p \\
p = 1, q = p^2 = 1^2 = 1.~~~\text{Point on the parabola is:}~~(1,1)
\]The point on the curve y = x^2 that has the minimum distance from the line y = 2x - 4 is (1,1).

Answer 2

\[\text{I meant }\frac{dS}{dp} \text{ and not } \frac{dS}{dq}.\]

Answer 3

Thanks a lot, you made it a lot more clear!