Second Degree Equations, xy nonzero, rotation of axes question

[Asphyxiated]

Homework Statement

x^{2}-4xy+y^{2} - 1= 0

Homework Equations

Rotation of axes equations:

x = X cos(\alpha)-Ysin(\alpha)

y = Xsin(\alpha) + Ycos(\alpha)

Alpha is given in the text as:

cot(2\alpha) = \frac {A-C}{B}

but I used the fact that:

cot(\alpha) = \frac {1}{tan(\alpha)}

to solve for alpha like so:

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

The Attempt at a Solution

ok so i try to find alpha first like:

\alpha = \frac {tan^{-1}(-\frac{4}{1-1})} {2}

which means that it is really just:

\alpha = \frac {tan^{-1}(0)} {2}

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:

x^{2}+2\sqrt{3}xy-y^{2}-7=0

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

2x^{2}-2y^{2}-7=0

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

 

[eumyang]

ok so i try to find alpha first like:

\alpha = \frac {tan^{-1}(-\frac{4}{1-1})} {2}

which means that it is really just:

\alpha = \frac {tan^{-1}(0)} {2}

It's not inverse tangent of 0. It's inverse tangent of -4/0. -4/0 is undefined.

I probably would have just worked with the cotangent equation:
\begin{aligned}<br /> cot(2\alpha) &amp;= \frac {A-C}{B} \\<br /> &amp;= \frac{1 - 1}{-4} \\<br /> &amp;= 0<br /> \end{aligned}

If 0 < α < π/2, then 0 < 2α < π, and the only place between 0 and π where the cotangent equals 0 is at π/2. So
\begin{aligned}<br /> 2\alpha &amp;= \frac {\pi}{2} \\<br /> \alpha &amp;= \frac{\pi}{4}<br /> \end{aligned}
... which is the angle you wanted.69