# Find the point on the curve y = x^2 that is closest to the point (0, b).

1. Nov 27, 2009

### plexus0208

1. The problem statement, all variables and given/known data
Find the point on the curve y = x2 that is closest to the point (0, b).

2. Relevant equations
Lagrange multipliers

3. The attempt at a solution
L(x,y,λ) = x2 + (y-b)2 - λ(x2-y)
Lx = 2x - 2λx = 0
Ly = 2y - 2b + λ = 0
Lλ = x2 - y = 0

Then solve for x and y? Is this right?

Last edited: Nov 27, 2009
2. Nov 27, 2009

### Dick

It's ok so far. Keep going.

3. Nov 27, 2009

### plexus0208

2x = 2λx -> λ = 1
2y = 2b + λ -> y = (2b+1)/2
x^2 = y -> x = ((2b+1)/2)^1/2

4. Nov 27, 2009

### Dick

You've got a sign error in your y equation. And you've also ignored a possible solution of the first equation. How about x=0? And in the last equation there are two solutions for x.

Last edited: Nov 27, 2009
5. Nov 27, 2009

### plexus0208

So basically, there are three points on the curve that are equally closest to (0,b)?

6. Nov 27, 2009

### Dick

Depends on b. The lagrange multiplier method only gives you the critical points. There might be one or three of them depending on the value of b. You then have to check them to figure out which is closest.

7. Nov 27, 2009

### plexus0208

Well, I was given the point (0, b), without any information about b, so I guess I should just put down all three points then...

8. Nov 27, 2009

### HallsofIvy

No, whatever b is you should check to see which points are minimum. You have only shown that they are critical points. It is quite possible that one of the points is a local maximum for the distance.

9. Nov 27, 2009

### plexus0208

Lxx = 2 - 2λ
Lyy = 2
Lxy = 0

if λ = 1, then discriminant = LxxLyy - Lxy2 = 0...

10. Nov 27, 2009

### Dick

Why don't you just plug the points you get into the distance formula and figure out which one is closest?