# Calc Optimization - Point on an ellipse closest to origin.

1. Nov 15, 2007

### jbot97

1. The problem statement, all variables and given/known data
x^2 - 2xy + 6y^2 = 10
Find the point on the ellipse closest to the origin (0,0).

2. Relevant equations

3. The attempt at a solution

Absolutely no one in my class can solve this. We've been to the math lab and none of the helpers there know how to solve it. I think the only person who knows how to solve it here is my professor, and he essentially dodges our requests for an example. I know how to solve this if the ellipse is, for example, x^3+4y^2 = 7. The problem is, this ellipse has 2xy in it and I don't know how to solve for y to plug in the distance formula.

For x^3 + 4y^2 = 10:

y = [(10-x^3)/4]^.5

D = sq rt {x^2 + [(10-x^3)^.5]^2}

When D is a min, you have the closest point to the origin.

So, I essentially just need to know how to solve x^2 - 2xy + 6y^2 = 10 for y, or an alternate means to solve this problem.

Thanks!

2. Nov 15, 2007

### Kreizhn

Complete the square with respect to y

3. Nov 15, 2007

### jbot97

Thanks

How do I do that?

4. Nov 15, 2007

### Kreizhn

It is a rather involved process. I assume you know how to complete the square for something like $ax^2+bx+c$. In this case treat x as a constant and complete the square for y

5. Nov 15, 2007

### jbot97

I understand how I could take the derivative of that with respect to y... but I have no idea how to go about completing the square like that...

6. Nov 15, 2007

### Kreizhn

$$6y^2-2xy+x^2 = 6 \left( y^2-\frac{1}{3} xy \right) + x^2$$
$$6 \left( y^2-\frac{1}{3} xy + \frac{1}{36} x^2 - \frac{1}{36} x^2 \right) + x^2$$
$$6 \left( y^2-\frac{1}{3} xy + \frac{1}{36} x^2 \right) - \frac{1}{6} x^2 + x^2$$
$$6 \left( y- \frac{1}{6} x \right)^2 +\frac{5}{6} x^2$$

Edit: Sorry I destroyed the Typesetting so bad, trying to do math in two variables and type set at the same time threw me off

Last edited: Nov 15, 2007
7. Nov 15, 2007

### jbot97

Thank you so much! I think I can figure this one out now.