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.