Equation of ellipse with knowing it’s centre and ends of conjugate diameters’

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The coordinate system is affine.

Write the equation of the ellipse, knowing it’s center $C (2, 1)$ and the ends of two conjugate diameters $A (5, 1)$, $B (0, 3)$.

The book proposes to use vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ as unit vectors of $O’x’$ and $O’y’$, but I don’t see how it’s helpful in the affine system.

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The parametric equation of the ellipse using its conjugate radii $\vec{CA}, \vec{CB}$ is

$ \vec{P}(t) = \vec{C} + \cos t \vec{CA} + \sin t \vec{CB} \hspace{48pt}(1)$

In matrix-vector form, let $P = [x, y]^T, C = [2,1]^T $ then $CA =[3, 0]^T $ and $CB = [-2, 2]^T $ and $u = [\cos t , \sin t ]^T$, the above equation becomes

$ P - C = G u \hspace{48pt} (2) $

where $G = \begin{bmatrix} 3 && - 2 \\ 0 && 2 \end{bmatrix} $

From equation $(2)$ we have

$ u = G^{-1} (P - C) \hspace{48pt} (3) $

And since $\cos^2 t + \sin^2 t = 1 $ , then $u^T u = 1 $ , hence,

$ (P - C)^T G^{-T} G^{-1} (P - C) = 1 \hspace{48pt} (4) $

And this is the equation of the ellipse. For numerical evaluation, we have

$G^{-1} = \dfrac{1}{6} \begin{bmatrix} 2 && 2 \\ 0 && 3 \end{bmatrix} $

Thus

$G^{-T} G^{-1} = \dfrac{1}{36} \begin{bmatrix} 4 && 4 \\ 4 && 13 \end{bmatrix} $

so that the equation of the ellipse $(4)$ becomes

$ 4 (x - 2)^2 + 8 (x - 2)(y - 1) + 13 (y - 1)^2 = 36 $