# Help with Rotation of Equations

• Aug 19th 2008, 03:38 PM
The Real Cosby
Help with Rotation of Equations
8x^2 - 4xy + 5y^2 = 36

@=theta

1. Find tan 2@ and find cos 2@

2. Find sin@ and cos@

3. Transform the equation xy=1 by rotation, thus eliminating the xy term.
• Aug 20th 2008, 09:02 AM
Soroban
Hello, The Real Cosby!

You should be familiar with the "Rotation" formulas.
Otherwise, you should not have been assigned this problem.

The general quadratic equation is: . $Ax^2 + Bxy + Cy^2 + Dx + Ey + F \;=\;0$

Quote:

$8x^2 - 4xy + 5y^2 \:=\: 36\;\;{\color{blue}[1]}$

1. Find $\tan2\theta$ and $\cos2\theta$

We have: . $A = 8,\;B = -4,\;C = 5,\;F = -36$

Formula: . $\tan2\theta \;=\;\frac{B}{A-C}$

We have: . $\tan2\theta \:=\:\frac{-4}{8-5} \:=\:\boxed{-\frac{4}{3}}$

Since $\theta$ is in Quadrant 2: . $\tan2\theta \:=\:\frac{4}{-3} \:=\:\frac{opp}{adj}$

$2\theta$ is in a right triangle with: $opp = 4,\;adj = -3$
Using Pythagorus, we get: . $hyp = 5$

Therefore: . $\cos2\theta \:=\:\frac{adj}{hyp} \:=\:\boxed{-\frac{3}{5}}$

Quote:

2. Find $\sin\theta$ and $\cos\theta$
Since $2\theta$ is in Quadrant 2, then $\theta$ is in Quadrant 1.

Identities: . $\begin{Bmatrix}\sin^2\theta &=&\dfrac{1-\cos2\theta}{2} \\ \\[-4mm] \cos^2\theta &=&\dfrac{1+\cos2\theta}{2} \end{Bmatrix}$

We have: . $\begin{array}{ccccccccc}\sin^2\theta &=& \dfrac{1 -(-\frac{3}{5})}{2} &=& \dfrac{4}{5}& \Rightarrow & \sin\theta &=&\boxed{\dfrac{2}{\sqrt{5}}} \\
\cos^2\theta &=& \dfrac{1 + (-\frac{3}{5})}{2} &=& \dfrac{1}{5} & \Rightarrow & \cos\theta &=&\boxed{\dfrac{1}{\sqrt{5}}}\end{array}$

Quote:

3. Transform the equation xy=1 by rotation, . ??
. . thus eliminating the xy term.

Formulas: . $\begin{Bmatrix}x' &=& x\cos\theta - y\sin\theta \\ y' &=& x\sin\theta + y\cos\theta \end{Bmatrix}$

We have: . $\begin{array}{ccccc}x' &=&\frac{1}{\sqrt{5}}x - \frac{2}{\sqrt{5}}y &=& \dfrac{x-2y}{\sqrt{5}} \\ \\[-4mm]
y' &=&\frac{2}{\sqrt{5}}x + \frac{1}{\sqrt{5}}y &=& \dfrac{2x+y}{\sqrt{5}} \end{array}$

Substitute into [1]: . $8\left(\frac{x-2y}{\sqrt{5}}\right)^2 - 4\left(\frac{x-2y}{\sqrt{5}}\right)\left(\frac{2x+y}{\sqrt{5}}\ri ght) + 5\left(\frac{2x+y}{\sqrt{5}}\right)^2 \;=\;36$

. . $\frac{8(x^2-4xy + 4y^2)}{5} - \frac{4(2x^2-3xy-2y^2)}{5} + \frac{5(4x^2+4xy + y^2)}{5} \;=\;36$

. . $\frac{8x^2 - 32xy + 32y^2 - 8x^2 + 12xy + 8y^2 + 20x^2 + 20xy + 5y^2}{5} \;=\;36$

. . $\frac{20x^2 + 45y^2}{5} \;=\;36 \quad\Rightarrow\quad 4x^2 + 9y^2 \;=\;36 \quad\Rightarrow\quad\boxed{ \frac{x^2}{9} + \frac{y^2}{4} \;=\;1}\quad\text{ an ellipse!}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Drat! .I just figured out what #3 meant . . .

It is a separate problem!

We have: . $xy \:=\:1\;\;{\color{blue}[2]}$
. . That is: . $A = 0,\;B = 1,\; C= 0,\;D = 0,\;E = 0,\;F = -1$

$\tan2\theta \:=\:\frac{1}{0-0} \:=\:\infty \quad\Rightarrow\quad 2\theta \:=\:\frac{\pi}{2} \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{4}$

. . It is a hyperbola rotated through 45°.

$\sin\theta = \frac{1}{\sqrt{2}},\;\cos\theta = \frac{1}{\sqrt{2}}$

Hence: . $\begin{Bmatrix}x' &=&\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y &=& \dfrac{x-y}{\sqrt{2}} \\ \\[-4mm]
y' &=&\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y &=& \dfrac{x+y}{\sqrt{2}} \end{Bmatrix}$

Substitute into [2]: . $\left(\frac{x-y}{\sqrt{2}}\right)\left(\frac{x+y}{\sqrt{2}}\righ t) \;=\;1 \quad\Rightarrow\quad \frac{x^2-y^2}{2} \;=\;1$

. . Therefore: . $\boxed{x^2-y^2 \;=\;2}$

I need a nap . . .
.
• Aug 20th 2008, 12:28 PM
The Real Cosby
Wow sir, you get a big thanks from me. (Clapping)

My mind was off and I was just doing some other stuff.

I think I need a nap too...