# Thread: Axis rotation with fractional angles

1. ## Axis rotation with fractional angles

Part of the algebra I'm doing with this is producing bad results, please tell me why.

Graph: $\displaystyle 11x^2+7xy-13y^2=621$
$\displaystyle arccos 24/7$ is a fraction so:
$\displaystyle sin\theta=\sqrt{\frac{1-\frac{24}{7}}{2}$$\displaystyle x=x'\frac{31}{14}-y'\frac{-17}{14} \displaystyle cos\theta=\sqrt{\frac{1+\frac{24}{7}}{2}$$\displaystyle y=x'\frac{-17}{14}+y;\frac{31}{14}$

I try to solve and eliminate x'y' but end up with odd fractions:

$\displaystyle 11[x'\frac{31}{14}+y'\frac{17}{14}]^2+7[x'\frac{31}{14}+y'\frac{17}{14}][x'\frac{-17}{14}+y'\frac{31}{14}]-13[x'\frac{-17}{14}+y\frac{31}{14}]^2=621$

~~~~~~~~~~~~~~~~~~~~~

Which when I calculate gives me:

$\displaystyle 11[x'^2\frac{961}{196}+2x'y'\frac{527}{196}+y'^2\frac {289}{196}]+7[x'^2\frac{527}{196}+x'y'\frac{961}{196}-x'y'\frac{289}{196}+y'^2\frac{527}{196}]-13[x'^2\frac{289}{196}-2x'y'\frac{527}{196}+y'^2\frac{961}{196}]=621$

Which can't be right because the x'y' will not equal 0.

2. ## Re: Axis rotation with fractional angles

Originally Posted by Greymalkin
Part of the algebra I'm doing with this is producing bad results, please tell me why.
Graph: $\displaystyle 11x^2+7xy-13y^2=621$
$\displaystyle arccos 24/7$ is a fraction so:
$\displaystyle sin\theta=\sqrt{\frac{1-\frac{24}{7}}{2}$$\displaystyle x=x'\frac{31}{14}-y'\frac{-17}{14} \displaystyle cos\theta=\sqrt{\frac{1+\frac{24}{7}}{2}$$\displaystyle y=x'\frac{-17}{14}+y;\frac{31}{14}$
What is the question? There does't not appear to be any question there.

3. ## Re: Axis rotation with fractional angles

Originally Posted by Plato
What is the question? There does't not appear to be any question there.
Graph$\displaystyle 11x^2+7xy-13y^2=621$

"Part of my algebra is producing bad results, please tell me where/why my calculations are producing x'y' terms that are irreducible.", with algebra below.

4. ## Re: Axis rotation with fractional angles

Seems that I found "an" error in $\displaystyle cos2\theta$ yet am still getting bad results with the algebra, am I doing anything illegal above??

Edit: Here is my new work with the above error taken into account,

$\displaystyle sin\theta=\sqrt{\frac{1-\frac{24}{25}}{2}$$\displaystyle =\frac{1}{50} --- \displaystyle x=x'\frac{49}{50}-y'\frac{1}{50} \displaystyle cos\theta=\sqrt{\frac{1+\frac{24}{25}}{2}$$\displaystyle =\frac{49}{50}$ --- $\displaystyle y=x'\frac{1}{50}+y'\frac{49}{50}$

Continuing on>>>

$\displaystyle 11x^2+7xy-13y^2=621$

$\displaystyle 11[x'\frac{49}{50}-y'\frac{1}{50}]^2+7[(x'\frac{49}{50}-y'\frac{1}{50})(x'\frac{1}{50}+y'\frac{49}{50}]-13[x'\frac{1}{50}+y'\frac{49}{50}]^2=621$

~~~~~~~~~~~~~~~~~~~~~~~~~

$\displaystyle 11[x'^2\frac{2401}{2500}-2x'y'\frac{49}{2500}+y'^2\frac{1}{2500}]+7[x'^2\frac{49}{2500}+x'y'\frac{2401}{2500}-x'y'\frac{1}{2500}-y'^2\frac{49}{2500}]-13[x'^2\frac{1}{2500}+2x'y'\frac{49}{2500}+y'^2\frac{ 2401}{2500}]=621$

After simplifying, something is wrong because the x'y' terms, yet again, will not dissapear for me.