# Find the critical points of (x^2+y^2-4)(x+y) and their nature

1. Aug 7, 2011

### rbnvrw

Hi,

I am studying for my Analysis final and came across this problem I just can't get my head around:
1. The problem statement, all variables and given/known data
Find the critical points of $f(x,y) = (x^2+y^2-4)(x+y)$ and their nature.

2. Relevant equations
$\vec{\nabla} f(x,y) = \vec{0}$

3. The attempt at a solution
$\frac{\partial f(x,y)}{\partial x} = 3x^2+2xy+y^2-4 = 0$
$\frac{\partial f(x,y)}{\partial y} = 3y^2+2xy+x^2-4 = 0$
These are both ellipses with center (0,0) and angles $\frac{5\pi}{8}$ and $\frac{\pi}{8}$ respectively.
According to Wolfram Alpha, there are four intersection points. (Click here to view) However, I need to solve this problem analytically and I wish to understand how it is done.
I have tried to eliminate a variable from the equations, but it is not possible due to the $2xy$ term. Another solution would be to write the ellipses in a parametric form, but this I feel would be overly complicated.

Could anyone please shed some light on this?

2. Aug 7, 2011

### HallsofIvy

Staff Emeritus
The first thing I would do is subtract one equation from the other to get
$$3(x^2- y^2)+ (y^2- x^2)= 2(x^2- y^2)= 0$$
so that either y= x or y= -x.

3. Aug 7, 2011

### ehild

Subtract the second equation from the first one and factorise.

ehild

4. Aug 7, 2011

### rbnvrw

However, this doesn't give me the four points of intersection. If I take the solutions $y=\pm x$, I get the solutions $(0,0) \wedge (\pm \sqrt{2},\pm \sqrt{2})$.
However, the points $(\pm \sqrt{\frac{2}{3}},\pm \sqrt{\frac{2}{3}})$ are also solutions of the equation. And the point $(0,0)$ is not a solution at all.
How would I get the other two points?

5. Aug 7, 2011

### ehild

Your values are not correct. Give the solutions as pairs.
(√2, √2) is not a solution, but (√2, -√2) is.
(0,0) is not a solution. How did you get it? If you substitute x=y into one of the original equation, you will get the pair with √(2/3)

ehild

Last edited: Aug 7, 2011
6. Aug 7, 2011

### rbnvrw

I got this:
$y = \pm x$
$(x^2+x^2-4)(x+x)=0$
$x(x^2-2)=0$
$x = 0 \vee x=\pm \sqrt{2}$
According to the solutions manual, which only states the values, not the method, the solutions are:
$x = \pm \sqrt{2} \vee x = \pm \sqrt{\frac{2}{3}}$

7. Aug 7, 2011

### rbnvrw

Oops... I see it now. I used f(x,y) instead of the partial derivatives.
I will try again!

8. Aug 7, 2011

### rbnvrw

Yes! I got it!
Thank you for your help, I appreciate it! :D

9. Aug 7, 2011

### ehild

You see, it was worth to write it again...

ehild