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

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

1. Find $\displaystyle \tan2\theta$ and $\displaystyle \cos2\theta$
We have: .$\displaystyle A = 8,\;B = -4,\;C = 5,\;F = -36$

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

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

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

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

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

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

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

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

3. Transform the equation xy=1 by rotation, . ??
. . thus eliminating the xy term.
Formulas: . $\displaystyle \begin{Bmatrix}x' &=& x\cos\theta - y\sin\theta \\ y' &=& x\sin\theta + y\cos\theta \end{Bmatrix}$

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

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

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

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

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

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

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

It is a separate problem!

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

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

. . It is a hyperbola rotated through 45°.

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

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

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

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

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...