- #1
themadhatter1
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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?