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

Hi,

I am studying for my Analysis final and came across this problem I just can't get my head around:

## Homework Statement

Find the critical points of $f(x,y) = (x^2+y^2-4)(x+y)$ and their nature.

## Homework Equations

$\vec{\nabla} f(x,y) = \vec{0}$

## 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?

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HallsofIvy
Homework Helper
Hi,

I am studying for my Analysis final and came across this problem I just can't get my head around:

## Homework Statement

Find the critical points of $f(x,y) = (x^2+y^2-4)(x+y)$ and their nature.

## Homework Equations

$\vec{\nabla} f(x,y) = \vec{0}$

## The Attempt at a Solution

$\frac{\partial f(x,y)}{\partial x} = 3x^2+2xy+y^2-4 = 0$
$\frac{\partial f(x,y)}{\partial y} = 3y^2+2xy+x^2-4 = 0$
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.

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?

ehild
Homework Helper
Subtract the second equation from the first one and factorise.

ehild

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?

ehild
Homework Helper
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:
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:
$x = \pm \sqrt{2} \vee x = \pm \sqrt{\frac{2}{3}}$

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

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

ehild
Homework Helper
You see, it was worth to write it again...

ehild