# Thread: Can someone help me convert these two?

1. ## Can someone help me convert these two?

Find the angle of rotation for the following conic that will eliminate the xy term. Rotate the axes through the angle of rotation, eliminate the xy term and write the conic in standard form.
xy=-10

Given the following point in polar form, change it to rectangular form.
(2,-(pi/3))

I'll be glad when this class is over!

2. Originally Posted by Dave53
Find the angle of rotation for the following conic that will eliminate the xy term. Rotate the axes through the angle of rotation, eliminate the xy term and write the conic in standard form.
xy=-10

Given the following point in polar form, change it to rectangular form.
(2,-(pi/3))

I'll be glad when this class is over!
For the second one we know that

$(x,y)\Rightarrow{(r\cos(\theta),r\sin(\theta)}$

so

$\left(2,-\frac{\pi}{3}\right)\Rightarrow\left(1,-\sqrt{3}\right)$

3. Originally Posted by Dave53
Find the angle of rotation for the following conic that will eliminate the xy term. Rotate the axes through the angle of rotation, eliminate the xy term and write the conic in standard form.
xy=-10

Given the following point in polar form, change it to rectangular form.
(2,-(pi/3))

I'll be glad when this class is over!
Well this has asymptotes perpendicular to the standard form so I would have guessed $\pi/4$. Having not done this in years, if did some googeling (see http://www.mrjohns.net/ihs/0708/clas...Rotations3.pdf p. 4) I found that

$\phi = \frac{1}{2}cot^{-1}\left(\frac{A-C}{B}\right)$ Where A, B, C are the $x^2,\ xy,\ y^2$ terms respectively.

So we get $\phi = \frac{1}{2}cot^{-1}\left(0\right)=\frac{1}{2}\frac{\pi}{2}=\frac{\p i}{4}$

So then let
$x=x'\cos \phi-y' \sin \phi$
$y=x'\sin \phi+y' \cos \phi$

Howzat?