Rotation Formulas

**magentarita** · October 18th 2008, 04:24 AM

Determine the appropriate rotation formulas to use so that the new equation contains no xy-term.

(1) x^2 - 4xy + y^2 - 3 = 0

(2) 3x^2 - 10xy + 3y^2 - 32 = 0

**Soroban** · October 18th 2008, 08:28 AM

Hello, magentarita!

I'll walk through the first one . . .

Determine the appropriate rotation formulas to use
so that the new equation contains no $xy$ -term.

Given the general quadratic: . $Ax^2 + Bxy + Cy^2 + Dx + Ey + F \:=\:0$

. . the angle of rotation $\theta$ is given by: . $\tan2\theta \:=\:\frac{B}{A-C}$

. . and the new coordinates are: . $\begin{Bmatrix}x &=& X\cos\theta - Y\sin\theta \\ y &=& X\sin\theta + Y\cos\theta \end{Bmatrix}$

$(1)\;\;x^2 - 4xy + y^2 \:=\: 3$

We have: . $A = 1,\;B = -4,\;C = 1$

. . Then: . $\tan2\theta \:=\:\frac{-4}{1-1} \:=\:-\infty \quad\Rightarrow\quad 2\theta\:=\:-\frac{\pi}{2}\quad\Rightarrow\quad \theta \:=\:-\frac{\pi}{4}$

. . Hence: . $\sin\theta \:=\: -\frac{1}{\sqrt{2}},\quad \cos\theta \:=\: \frac{1}{\sqrt{2}}$

And we have: . $\begin{array}{ccccc}x &=& X\left(\frac{1}{\sqrt{2}}\right) - Y\left(\text{-}\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(X + Y) \\ \\[-3mm]<br /> y&=& X\left(\text{-}\frac{1}{\sqrt{2}}\right) + Y\left(\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(\text{-}X + Y) \end{array}$

Substitute into: . $x^2-4xy + y^2 \;=\;3$

. . $\bigg[\frac{1}{\sqrt{2}}(X + Y)\bigg]^2 - 4\bigg[\frac{1}{\sqrt{2}}(X + Y)\bigg]\bigg[\frac{1}{\sqrt{2}}(\text{-}X+Y)\bigg] + \bigg[\frac{1}{\sqrt{2}}(\text{-}X + Y)\bigg]^2 \;=\;3$

. . $\tfrac{1}{2}(X^2+2XY + Y^2) - 2(\text{-}X^2+Y^2) + \tfrac{1}{2}(X^2-2XY + Y^2) \;=\;3$

. . $\tfrac{1}{2}X^2 + XY + \tfrac{1}{2}Y^2 + 2X^2 - 2Y^2 + \tfrac{1}{2}X^2 - XY + \tfrac{1}{2}Y^2 \;=\;3$

. . $3X^2 - Y^2 \;=\;3$

. . $\frac{X^2}{1} - \frac{Y^2}{3} \;=\;1 \quad\hdots\quad \text{hyperbola}$

**magentarita** · October 18th 2008, 09:14 PM

Originally Posted by Soroban

Hello, magentarita!

I'll walk through the first one . . .

Given the general quadratic: . $Ax^2 + Bxy + Cy^2 + Dx + Ey + F \:=\:0$

. . the angle of rotation $\theta$ is given by: . $\tan2\theta \:=\:\frac{B}{A-C}$

. . and the new coordinates are: . $\begin{Bmatrix}x &=& X\cos\theta - Y\sin\theta \\ y &=& X\sin\theta + Y\cos\theta \end{Bmatrix}$

We have: . $A = 1,\;B = -4,\;C = 1$

. . Then: . $\tan2\theta \:=\:\frac{-4}{1-1} \:=\:-\infty \quad\Rightarrow\quad 2\theta\:=\:-\frac{\pi}{2}\quad\Rightarrow\quad \theta \:=\:-\frac{\pi}{4}$

. . Hence: . $\sin\theta \:=\: -\frac{1}{\sqrt{2}},\quad \cos\theta \:=\: \frac{1}{\sqrt{2}}$

And we have: . $\begin{array}{ccccc}x &=& X\left(\frac{1}{\sqrt{2}}\right) - Y\left(\text{-}\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(X + Y) \\ \\[-3mm]$ $<br /> <font size=$ y&=& X\left(\text{-}\frac{1}{\sqrt{2}}\right) + Y\left(\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(\text{-}X + Y) \end{array}" alt="
y&=& X\left(\text{-}\frac{1}{\sqrt{2}}\right) + Y\left(\frac{1}{\sqrt{2}}\right) &=& \frac{1}{\sqrt{2}}(\text{-}X + Y) \end{array}" />

Substitute into: . $x^2-4xy + y^2 \;=\;3$

. . $\bigg[\frac{1}{\sqrt{2}}(X + Y)\bigg]^2 - 4\bigg[\frac{1}{\sqrt{2}}(X + Y)\bigg]\bigg[\frac{1}{\sqrt{2}}(\text{-}X+Y)\bigg] + \bigg[\frac{1}{\sqrt{2}}(\text{-}X + Y)\bigg]^2 \;=\;3$

. . $\tfrac{1}{2}(X^2+2XY + Y^2) - 2(\text{-}X^2+Y^2) + \tfrac{1}{2}(X^2-2XY + Y^2) \;=\;3$

. . $\tfrac{1}{2}X^2 + XY + \tfrac{1}{2}Y^2 + 2X^2 - 2Y^2 + \tfrac{1}{2}X^2 - XY + \tfrac{1}{2}Y^2 \;=\;3$

. . $3X^2 - Y^2 \;=\;3$

. . $\frac{X^2}{1} - \frac{Y^2}{3} \;=\;1 \quad\hdots\quad \text{hyperbola}$

Thank you for taking time to show me step by step.
This is what I need most of the time.

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