• Apr 14th 2010, 11:53 PM
VitaX
$x^2 + 2\sqrt{3}xy + 3y^2 + 8\sqrt{3}x - 8y + 32 = 0$

A) Use the Discriminant test to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola.

B) Find a suitable rotation of axis

C) Find an equation for the graph in a $x'y'$ - plane

D) Sketch the graph over top both coordinate systems with your angle of rotation displayed.

My Work:

A) $B^2 - 4AC \rightarrow (2\sqrt{3})^2 - 4(1)(3) = 0$ So, The equation is a parabola.

B) Here's where I get somewhat confused, so bear with my work.
$tan 2\alpha = \frac{B}{A-C} \rightarrow \frac{2 \sqrt{3}}{1-3} \rightarrow \frac{\sqrt{3}}{-1}$
$2 \alpha = tan^{-1} \left(\frac{-\pi}{3}\right)$
$\alpha = \frac{-\pi}{6}$

From here on I'm not really quite sure what to do. The notes I have on this don't really give me enough help to continue on. I know later on though I have to utilize these:

$A'=Acos^2 \alpha + Bcos\alpha sin\alpha + Csin^2 \alpha$

$B'=Bcos2\alpha + (C-A)sin2 \alpha$

$C'=Asin^2 \alpha - Bcos\alpha sin\alpha + Ccos^2 \alpha$

$D'=Dcos\alpha + Esin\alpha$

$E'=-Dsin\alpha + Ecos\alpha$

$F'=F$
• Apr 15th 2010, 12:13 AM
HallsofIvy
Since you know that $\alpha= -\frac{\pi}{6}$, you can immediately calculate $cos(\alpha)=\frac{\sqrt{3}}{2}$ and $sin(\alpha)= -\frac{1}{2}$ and put them into those equations. What is stopping you?
• Apr 15th 2010, 12:24 AM
VitaX
Quote:

Originally Posted by HallsofIvy
Since you know that $\alpha= -\frac{\pi}{6}$, you can immediately calculate $cos(\alpha)=\frac{\sqrt{3}}{2}$ and $sin(\alpha)= -\frac{1}{2}$ and put them into those equations. What is stopping you?

So I just plug in those values into the equations I stated above and come up with the values for A',B',C',D',E' and come up with the new "shifted" equation?
• Apr 15th 2010, 12:42 AM
HallsofIvy
That's what the formulas say isn't it?
• Apr 15th 2010, 02:42 AM
VitaX
$A'=1\left(\frac{\sqrt{3}}{2}\right)^2 + 2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\left(\fra c{-1}{2}\right) + 3\left(\frac{-1}{2}\right)^2 \rightarrow A'=\frac{3}{4}$
$B'=2 \sqrt{3}\left(\frac{1}{2}\right) + (3-1)(\frac{-\sqrt{3}}{2} \rightarrow B'=0$
$C'=1\left(\frac{-1}{2}\right)^2 - 2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\left(\fra c{-1}{2}\right) + 3\left(\frac{\sqrt{3}}{2}\right)^2 \rightarrow C'=\frac{19}{4}$
$D'=8\sqrt{3}\left(\frac{\sqrt{3}}{2}\right) + (-8)\left(\frac{-1}{2}\right) \rightarrow D'=16$
$E'=-8\sqrt{3}\left(\frac{-1}{2}\right) + (-8)\left(\frac{\sqrt{3}}{2}\right) \rightarrow E'=0$
$F'=32$

$\frac{3}{4}(x')^2 + \frac{19}{4}(y')^2 + 16x' + 32 = 0$

Is this equation correct for the graph in the $x'y'$ - plane? I'm having a little trouble putting this in the correct form of a parabola (if this equation is correct that is), if someone could point me in the right direction that would be great.
• Apr 15th 2010, 03:06 PM
VitaX
Am I supposed to utilize $x=x'cos\alpha - y'sin\alpha$ and $y=x'sin\alpha + y'cos\alpha$
I'm so confused right now. (Headbang)