(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex] x^{2}-4xy+y^{2} - 1= 0 [/tex]

2. Relevant equations

Rotation of axes equations:

[tex] x = X cos(\alpha)-Ysin(\alpha) [/tex]

[tex] y = Xsin(\alpha) + Ycos(\alpha) [/tex]

Alpha is given in the text as:

[tex] cot(2\alpha) = \frac {A-C}{B} [/tex]

but I used the fact that:

[tex] cot(\alpha) = \frac {1}{tan(\alpha)} [/tex]

to solve for alpha like so:

[tex] \alpha = \frac {tan^{-1}(\frac{B}{A-C})}{2} [/tex]

3. The attempt at a solution

ok so i try to find alpha first like:

[tex] \alpha = \frac {tan^{-1}(-\frac{4}{1-1})} {2} [/tex]

which means that it is really just:

[tex] \alpha = \frac {tan^{-1}(0)} {2} [/tex]

and the inverse tangent of 0 is 0 degrees, the only problem is that I am looking for 45 degrees, where did I go wrong? Is there something wrong with me using the trig relationship to solve for alpha?

thanks!

EDIT____________________________________

I just used the same alpha equation with an inverse tangent on the equation:

[tex] x^{2}+2\sqrt{3}xy-y^{2}-7=0 [/tex]

and got alpha at 30 degrees and a simplified equation of:

[tex] 2x^{2}-2y^{2}-7=0 [/tex]

which all passes discrimination tests so what exactly is wrong with the first problem? of course this only worked because A-C was nonzero...

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# Homework Help: Second Degree Equations, xy nonzero, rotation of axes question

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