I would like to explain the transforms on a picture.

$\frac{x'^2}{a^2}+\frac{y'^2}{b^2}=1$
$x=x'.\cos \alpha - y'. \sin \alpha $
$y=x'.\sin \alpha + y'. \cos \alpha $
We can get easily the result above from the picture
$\cos \alpha. x=\cos \alpha (x'.\cos \alpha - y'. \sin \alpha) $
$\sin \alpha. y=\sin \alpha (x'.\sin \alpha + y'. \cos \alpha) $
$x'=x.\cos \alpha + y. \sin \alpha $
$-\sin \alpha. x=\cos \alpha (x'.\cos \alpha - y'. \sin \alpha) $
$\cos \alpha. y=\sin \alpha (x'.\sin \alpha + y'. \cos \alpha) $
$y'=-x.\sin \alpha + y. \cos \alpha $
$\frac{(x.\cos \alpha + y. \sin \alpha)^2}{a^2}+\frac{(-x.\sin \alpha + y. \cos \alpha)^2}{b^2}=1$
$(\frac{\cos^2 \alpha }{a^2}+\frac{\sin^2 \alpha }{b^2})x^2+2(\frac{\cos \alpha \sin \alpha }{a^2}-\frac{\cos \alpha \sin \alpha }{b^2})xy+(\frac{\sin^2 \alpha }{a^2}+\frac{\cos^2 \alpha }{b^2})y^2=1$
$(\frac{\cos^2 \alpha }{a^2}+\frac{\sin^2 \alpha }{b^2})x^2+ \sin 2\alpha (\frac{1 }{a^2}-\frac{1}{b^2})xy+(\frac{\sin^2 \alpha }{a^2}+\frac{\cos^2 \alpha }{b^2})y^2=1$
$Ax^2+ Bxy+Cy^2=1$
$x=X-h$
$y=Y-k$
$A(X-h)^2+ B(X-h)(Y-k)+C(Y-k)^2=1$
$A=\frac{\cos^2 \alpha }{a^2}+\frac{\sin^2 \alpha }{b^2}$
$B=\sin 2\alpha (\frac{1 }{a^2}-\frac{1}{b^2})$
$C=\frac{\sin^2 \alpha }{a^2}+\frac{\cos^2 \alpha }{b^2}$