# Thread: Help with Rotation of Equations

1. ## 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.

2. 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: . $Ax^2 + Bxy + Cy^2 + Dx + Ey + F \;=\;0$

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

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

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

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 . . .
.

3. Wow sir, you get a big thanks from me.

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

I think I need a nap too...