Eliminating XY-Term: Solving for Rotated Axes in Conic Sections

[TonyC]

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.

PLEASE HELP!

 

[TD]

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

 

[arildno]

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)

 

[TonyC]

Can you show me how?

 

[arildno]

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]

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]

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 θ so that the "uv" term has 0 coefficient.

 

[saltydog]

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)

 

[TonyC]

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