I need to graph:

(x^2)+3xy+(y^2)=1

but, using the cot(2theta)=(a-c)/b

I got 0, and using

cot = 0degrees

or

cot = 90degrees

does not remove the xy term.

please help!!!

- May 23rd 2009, 11:59 PMmathmanaRotation of Axes to graph a quadratic equation with a xy term
I need to graph:

(x^2)+3xy+(y^2)=1

but, using the cot(2theta)=(a-c)/b

I got 0, and using

cot = 0degrees

or

cot = 90degrees

does not remove the xy term.

please help!!! - May 24th 2009, 01:57 AMmr fantastic
- May 24th 2009, 09:42 AMmathmana
Yes, I figured that out yesterday.

For some reason I thought cot was y/x.

Thanks for responding anyway!!