Rotate Hyperbola: Sketch Graph of xy-2y-4x=0

[themadhatter1]

Homework Statement

Rotate the axis to eliminate the xy-term. Sketch the graph of the equation showing both sets of axis.

xy-2y-4x=0

Homework Equations

[tex]\cot2\theta=\frac{A-C}{B}[/tex]
[tex]x=x'\cos\theta-y'\sin\theta[/tex]
[tex]y=x'\sin\theta+y'\cos\theta[/tex]

The Attempt at a Solution

xy-2y-4x=0

First I find the angle of the x' y' axis.
using
[tex]\cot2\theta=\frac{A-C}{B}[/tex]

I find it to be[tex]\theta=\frac{\pi}{4}[/tex]

Then find the x' and y' components

by using the second 2 equations I listed I come out with.
[tex]x=\frac{x'-y'}{\sqrt{2}}[/tex]
[tex]y=\frac{x'+y'}{\sqrt{2}}[/tex]Then comes the substitutions and simplifying.

[tex](\frac{x'-y'}{\sqrt{2}})(\frac{x'+y'}{\sqrt{2}})-2(\frac{x'+y'}{\sqrt{2}})-4(\frac{x'-y'}{\sqrt{2}})=0[/tex]

[tex]\frac{x'^2+y'^2}{2}+\frac{-6x'+2y'}{\sqrt{2}}=0[/tex]

[tex]\sqrt{2}(x')^2+\sqrt{2}(y')^2-12x'+4y'=0[/tex]

I complete the square and end up with

[tex](x'-\frac{12}{\sqrt{2}})^2+(y'+\frac{2}{\sqrt{2}})^2=20\sqrt{2}[/tex]

then I would divide though to get a 1 on the RHS but this is wrong.

The answer should be.

[tex] \frac{(x'-3\sqrt{2})^2}{16}-\frac{(y'-\sqrt{2})^2}{16}=1[/tex]

where did I go wrong?

 

[themadhatter1]

Never mind. I found my problem.

 

[ehild]

Then comes the substitutions and simplifying.

[tex](\frac{x'-y'}{\sqrt{2}})(\frac{x'+y'}{\sqrt{2}})-2(\frac{x'+y'}{\sqrt{2}})-4(\frac{x'-y'}{\sqrt{2}})=0[/tex]

It is correct up to here, but you mixed + and - after, and you made a mistake when multiplying the equation with 2. ehild