Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4

In essence I'm trying to take a horizontal ellipse and put it at an angle, rotating it around the origin. It's for my conics project, I'm making the elliptical orbits of the planets.

So I have the a, b, and c terms of the ellipse, (as in the distance from vertice to center and such) which are 28, 21, and $\displaystyle 7\sqrt{7}$ respectively. When plugged into the equation they give:

$\displaystyle {x}^{2} / 784 +{y}^{2} /441 =1$

which turns into

$\displaystyle 441 {x}^{2} + 784{y}^{2 } - 345,744 = 0$

from staring at my math book i've assumed that these x's and y's are actually x prime and y prime in the grand scheme of how we normally look at rotations.

so actually

$\displaystyle 441{x'}^{2 } + 784{y'}^{2 } - 345,744 = 0$

with this is mind I tried to apply the transformation equations knowing that I was trying to acheive a rotation of $\displaystyle \pi / 4$

$\displaystyle x = x' \cos \pi /4 - y' \sin \pi /4$

$\displaystyle x = (x' - y')/\sqrt{2}$

$\displaystyle \sqrt{2}x = x' - y'$

$\displaystyle \sqrt{2}y = x' + y'$

But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable and create a Bxy term.

So please, its due Wednesday... help?