Axis rotation with fractional angles

**Greymalkin** · September 5th 2012, 01:00 AM

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

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

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

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

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

Which when I calculate gives me:

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

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

**Plato** · September 5th 2012, 02:08 AM

Originally Posted by Greymalkin

Part of the algebra I'm doing with this is producing bad results, please tell me why.
Graph: $11x^2+7xy-13y^2=621$
$arccos 24/7$ is a fraction so:
$sin\theta=\sqrt{\frac{1-\frac{24}{7}}{2}$ $x=x'\frac{31}{14}-y'\frac{-17}{14}$
$cos\theta=\sqrt{\frac{1+\frac{24}{7}}{2}$ $y=x'\frac{-17}{14}+y;\frac{31}{14}$

What is the question? There does't not appear to be any question there.

**Greymalkin** · September 5th 2012, 02:25 AM

Originally Posted by Plato

What is the question? There does't not appear to be any question there.

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

"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.

**Greymalkin** · September 5th 2012, 02:57 AM

Seems that I found "an" error in $cos2\theta$ 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,

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

Continuing on>>>

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

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

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

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

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

**Greymalkin** · September 6th 2012, 02:08 AM

Bump- Still haven't solved this yet, please help!

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