Conic equations and rotations

**Såxon** · May 2nd 2009, 08:04 PM

Help with these two problems would be great:

Find the angle of revolution in degrees needed to put the conic in standard form. 2x^2-xy-3y^2-2x+4y-6=0
My work:

Modeling Cot2A= (A-C)/B
Cot2A= (2-(-)3)/-1 so Cot2A= (2+3)/-1 = 5/-1 = -5
I am very confused on how to go from here get the angle in degrees from a -5, any help please?
I know this isn't right but here goes,
Cot2A=-5 so Cot A = -25 degrees???

Find the equation for the conic in standard form and identify the conic.
x= 5 + 2 Cos t, y= -2 + sin t, 0 ≤ t ≤ 2π.

Do not really know what this means or where to start...
Thanks a lot for help!

**earboth** · May 2nd 2009, 09:39 PM

Originally Posted by Såxon

Help ... would be great:

...

Find the equation for the conic in standard form and identify the conic.
x= 5 + 2 Cos t, y= -2 + sin t, 0 ≤ t ≤ 2π.

Do not really know what this means or where to start...
Thanks a lot for help!

Eliminate the parameter t in the system of equations:

$\left|\begin{array}{l}x=5+2\cos(t)\\y=-2+\sin(t)\end{array}\right.$ $~\implies~$ $\left|\begin{array}{l}\dfrac12(x-5)=\cos(t)\\y+2=\sin(t)\end{array}\right.$

Now square both equations and add: (Keep in mind that $\sin^2(t) + \cos^2(t) = 1$ )

$\left|\begin{array}{l}\dfrac14(x-5)^2=\cos^2(t)\\(y+2)^2=\sin^2(t)\end{array}\right .$ $~\implies~$
$\dfrac{(x-5)^2}4 + \dfrac{(y+2)^2}1=1$

which is the equation of an allipse.

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