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:
$$\begin{aligned} cot(2\alpha) &= \frac {A-C}{B} \\ &= \frac{1 - 1}{-4} \\ &= 0 \end{aligned}$$

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
$$\begin{aligned} 2\alpha &= \frac {\pi}{2} \\ \alpha &= \frac{\pi}{4} \end{aligned}$$
... which is the angle you wanted.69