# Rotation and Systems of Quadratic Equations

• Jun 4th 2007, 06:16 PM
Rotation and Systems of Quadratic Equations
x^2-10xy+y^2+1=0

I'm supposed to rotate the axes in order to eliminate the xy-term. I'm told this problem is very simple, but I don't understand this section at all. Any help is greatly appreciated.
• Jun 4th 2007, 06:37 PM
ThePerfectHacker
Quote:

x^2-10xy+y^2+1=0

I'm supposed to rotate the axes in order to eliminate the xy-term. I'm told this problem is very simple, but I don't understand this section at all. Any help is greatly appreciated.

Given a general conic:
$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
If $B\not = 0$ (that means it has a cross term).

Then the required angle is:
$\cot 2\theta = \frac{A - C}{B}$
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You problem is,
$x^2 - 10xy+y^2 +1 =0$
The required angle satisfies,
$\cot 2\theta = \frac{ 1 -1 }{10} = 0$
Thus,
$2\theta = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$
---
When we rotate by $\theta$ we get the new coordinates:
$x'= x\cos \theta - y\sin \theta$
$y'= x\sin \theta + y\cos \theta$
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In this case $\theta = \frac{\pi}{4}$ thus,
$x' = x\cos \frac{\pi}{4} - y\sin \frac{\pi}{4} = x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2}$
$y'= x\sin \frac{\pi}{4} + y\cos \frac{\pi}{4} = x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2}$
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Substitute the new x and y coordinates:
$\left( x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2} \right)^2 - 10 \left(x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2}\right)\left( x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2} \right) + \left( x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2} \right)^2+1=0$

Expand,
$\left( \frac{1}{2}x^2 - xy + \frac{1}{2}y^2 \right) - 10 \left(\frac{1}{2}x^2 - \frac{1}{2}y^2 \right) + \left( \frac{1}{2}x^2+xy+\frac{1}{2}y^2 \right)+1=0$

Combine,
$-4x^2+6y^2+1=0$
• Jun 4th 2007, 07:14 PM
Thanks for helping me with the last problem. I know how to solve the rest of my homework except for this one:

xy-2y-4x=0

Am I supposed to change the equation first?
• Jun 4th 2007, 07:16 PM
ThePerfectHacker
Quote:

Hint: Think of this as $0x^2+1xy+0y^2 - 4x - 2y +0 =0$.