xy(x^2 - y^2) = C

**BenKiely** · September 11th 2012, 04:42 AM

I wanted to plot $xy(x^2-y^2)=C$ for C=840 but realised I couldn't isolate either x or y
I went to quickmath.com and entered the general equation and got...

Solve Equation :: QuickMath.com - Automatic Math Solutions

This returned 3 equations for calculating y; 2 complex and 1 real.

I know that for $xy(x^2-y^2)=840$ (7, 3), (7, 5) and (8, 7) are co-ordinates on the curve.
I cannot get these results by plugging in C=840, x=7 to the equations returned by quickmath.com.

Am I doing something wrong?

Thx

Ben

**pomp** · September 11th 2012, 04:59 AM

Hi Ben.

That is certainly a messy equation when you try to rearrange for y. Personally I would change coordinates.
Changing to polar: $x = r \cos \theta \,\ y = r \sin \theta$ , we get.

$r \cos \theta \cdot r \sin \theta \left( r^2 \cos^2 \theta - r^2 \sin^2 \theta \right) = C$

$\frac{1}{2} r^4 \sin 2 \theta \cos 2 \theta = C$

$\frac{r^4 }{4} \sin 4 \theta = C \,\ \rightarrow \,\ r^4 \sin 4 \theta = 4C$

Maybe it's easier to work with from here.

**BenKiely** · September 11th 2012, 07:36 AM

Originally Posted by pomp

Hi Ben.

Maybe it's easier to work with from here.

Thx pomp, yes, it's easier to do it this way from a graphing point of view.

**Wilmer** · September 11th 2012, 10:20 AM

The way you have it, 12 solutions:
x , y
-8,-7
-7,-5
-7,-3
-7,8
-5,7
-3,7
3,-7
5,-7
7,-8
7,3
7,5
8,7

**BenKiely** · September 11th 2012, 11:04 AM

That's right Wilmer, the equation produces 4 hyperbolae, one in each quadrant. My interest is in the top-right quadrant. i.e. where x and y are both positive.

**Wilmer** · September 11th 2012, 11:42 AM

Capish!

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Re: xy(x^2 - y^2) = C

Re: xy(x^2 - y^2) = C

Re: xy(x^2 - y^2) = C

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