# Thread: Rotation of conics (Hyperbola)

1. ## Rotation of conics (Hyperbola)

Hey I am really stuck on this problem:

2x^2 - 3xy - 2y^2 + 10 = 0

I know that when AC<0 then the rotation is for a Hyperbola, 2(-2)=-4
Now the formula for finding the degree of axis rotation is cot 2q=(A-C)/B
so (2-(-2))/-3 = -4/3, and this is where I am stuck, I can do the rest after I have q but whatever i try does not seem to be working
I need to solve cot q = -4/3

2. Originally Posted by fordyingseasons
...
Now the formula for finding the degree of axis rotation is cot 2q=(A-C)/B
so (2-(-2))/-3 = -4/3, and this is where I am stuck, I can do the rest after I have q but whatever i try does not seem to be working
I need to solve cot q = -4/3
Is this correct?
According to the formula you have

$\cot(2q) = -\frac43~\implies~ \tan(2q)=-\frac34~\implies~ 2q=-36.87^\circ = 143.13^\circ$

And now you can calculate q.

3. Thank you, i have another problem now with that answer however, I have to plug in xcosA - ysinB and xsinB + ycosA

In my book the give the example xcos(pie/4) - ysin(pie/4) = x(1/sqrt2)-y(1/sqrt2).
How do i plug in 71.565?