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

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

The Attempt at a Solution

xy-2y-4x=0

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

I find it to be\theta=\frac{\pi}{4}

Then find the x' and y' components

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

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

\frac{x'^2+y'^2}{2}+\frac{-6x'+2y'}{\sqrt{2}}=0

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

I complete the square and end up with

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

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

The answer should be.

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

where did I go wrong?

 

[themadhatter1]

Never mind. I found my problem.

 

[ehild]

Then comes the substitutions and simplifying.

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

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