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So lets say we have an ellipse with axes a and b and the rotation angle $\phi$ and center at $(0,0)$. Now I apply the rotation to $x^2/a^2+y^2/b^2=1$ getting

$$x' = x\cos(\phi) + y\sin(\phi)$$ $$y' = y\cos(\phi) + x\sin(\phi)$$

$$x^2(b^2cos(\phi)^2 + a^2sin(\phi)^2) + y^2(b^2\sin(\phi)^2 + a^2\cos(\phi)^2) + (b^2-a^2)\sin(\phi)\cos(\phi)xy - a^2b^2$$

This is some kind of quadratic form but I need to derive the quadratic form which I'll be able to convert to ellipse matrix.

What are the next steps to do this?

Thanks

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In general, an ellipse in a general position $[h,k]$ (what I needed) is implicitly given as $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$ Rotating the $x$ and $y$ coords yields (you're missing a $-$ sign in the first line) $$x' = x\cos(\phi) - y\sin(\phi)$$ $$y' = x\sin(\phi) + y\cos(\phi)$$ Plugging this into ellipse equation, you get $$\frac{([x\cos(\phi) - y\sin(\phi)]-h)^2}{a^2} + \frac{([x\sin(\phi) + y\cos(\phi)]-k)^2}{b^2} = 1$$ Now, you will need this in a $Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = 0$ form, as the ellipse matrix constitutes of the coefficients $A,B,C,D,E,$ and $F$. So you rewrite the previous, getting $$-(a^2 b^2) + b^2 h^2 + a^2 k^2 - 2 b^2 h x \cos(\phi) - 2 a^2 k y \cos(\phi) + b^2 x^2 \cos(\phi)^2 + a^2 y^2 \cos(\phi)^2 - 2 a^2 k x \sin(\phi) + 2 b^2 h y \sin(\phi) + 2 a^2 x y \cos(\phi) \sin(\phi) - 2 b^2 x y \cos(\phi) \sin(\phi) + a^2 x^2 \sin(\phi)^2 + b^2 y^2 \sin(\phi)^2 = 0$$ The coefficients therefore are $$\begin{array} AA &=& b^2 \cos^2(\phi) + a^2 \sin^2(\phi) \\ B &=& a^2 \cos(\phi) \sin(\phi) - b^2 \cos(\phi) \sin(\phi)\\ C &=& a^2 \cos^2(\phi) + b^2 \sin^2(\phi)\\ D &=& -b^2 h \cos(\phi) - a^2 k \sin(\phi)\\ E &=& -a^2 k \cos(\phi) + b^2 h \sin(\phi)\\ F &=& -a^2 b^2 + b^2 h^2 + a^2 k^2\end{array}$$ and the final ellipse matrix $$M = \begin{bmatrix}A & B & D \\ B & C & E \\ D & E & F \end{bmatrix}$$ By multiplying $\begin{bmatrix}x&y&1\end{bmatrix}M\begin{bmatrix}x&y&1\end{bmatrix}^T$, you get the original $Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = 0$ equation, hence the matrix form.

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