# Rotating axes to elimnate xy

I am having trouble rotating the axes to eliminate the xy-term.
8x^2+64xy+8y^2+12x+12y+9=0

I know Ax^2+Bxy+Cy^2+Dx+Ey+F=0
however, the professor is looking for an equation not an answer.

Here are my choices and I am stumped:
40x^2 + 12 sq rt2 x + 12 sq rt 2 y + 9 = 0

40x^2 - 24y^2 +12 sq rt2y +9 = 0

-24y^2 + 12 sq rt2x + 12 sq rt 2y + 9 = 0

40x^2 - 24y^ + 12 sq rt2x + 9 = 0

I am not sure what the process is to get to this point. I can run the basic Rotation Theorem for Conics and answer the problem with an angle degree.

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TD
Homework Helper
You can also reduce these conics to their standard Euclidean form using lineair algebra, by diagonalizing the corresponding matrix.

arildno
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Introduce new variables as follows:
$$u=x\cos\theta+y\sin\theta, v=-x\sin\theta+y\cos\theta$$
Here, $$\theta$$ is the angle between the positive x-axis and the positive u-axis.

Solve for x and y in terms of u and v, and gain:
$$x=u\cos\theta-v\sin\theta, y=u\sin\theta+v\cos\theta$$

Substitute these expressions for x and y in your equation, and eliminate the uv-term by making a smart choice of $$\theta$$

then, redefine "u" and "v" to be new "x" and "y".
(I don't see much point in doing so, but it seems your professor wants you to do that)

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Can you show me how?

arildno
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Put the expressions in u and v for x into the x's place, and similar for the expression for y.

For example, $$x^{2}=(u\cos\theta-v\sin\theta)^{2}=u^{2}\cos^{2}\theta-uv\sin(2\theta)+v^{2}\sin^{2}\theta$$

arildno
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If you haven't done so already, we also have:
$$xy=(u\cos\theta-v\sin\theta)(u\sin\theta+v\cos\theta)=\frac{u^{2}-v^{2}}{2}\sin(2\theta)+uv\cos(2\theta)$$
$$y^{2}=(u\sin\theta+v\cos\theta)^{2}=u^{2}\sin^{2}\theta+uv\sin(2\theta)+v^{2}\cos^{2}\theta$$
Use these expressions to your heart's content..

HallsofIvy
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You were given $$x=u\cos\theta-v\sin\theta, y=u\sin\theta+v\cos\theta$$
Go ahead and replace x and y in your equation by those, do the algebra and then pick &theta; so that the "uv" term has 0 coefficient.

saltydog
Homework Helper
Tony, what happen with all this? You figured the angle via:

$$Cot[2\alpha]=\frac{A-C}{B}$$

so you get:

$$\alpha=\frac{\pi}{4}$$

Great. Make the substitution:

$$x=\overline{x}Cos[\alpha]-\overline{y}Sin[\alpha]$$

$$y=\overline{x}Sin[\alpha]+\overline{y}Cos[\alpha]$$

Substitute those expressions into the equation right. For example the $8x^2$ would be:

$$8\left[\frac{\overline{x}}{\sqrt{2}}-\frac{\overline{y}}{\sqrt{2}}\right]^2$$

You can do the rest. Simplify to:

$$40\overline{x}^2-24\overline{y}^2+12\sqrt{2}\overline{x}+9=0$$

The $\overline{x}-\overline{y}$ axes are just 45 degrees tilted from the x-y axes. The equation is a hyperbola.

Right?

Edit: Corrected angle formula: it's Cot(2a)

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Thank you very much. This distance learning is extremely difficult. I am a visual person and your help is truly appreciated.