Project parabola onto another plane (edited) I am starting with a paraboloid in the form
$$A x^{2} + B x y + C y^{2} + D x + E y + F = z$$
and a plane
$$a x + b y + c z + d = 0$$
By equating the z terms in the two equations, I have the equation of the resulting 2nd degree curve in the world coordinate frame as
$$
f = \begin{cases} A_1 x^{2} + B_1 x y + C_1 y^{2} + D_1 x + E_1 y + F_1 = 0 \\
          z = -\frac{a x + b y + d}{c}
\end{cases}
$$
The first part of "f" is the projection of the curve of intersection along the z-axis onto the x-y plane.
For the case where this curve "f" is a parabola, I am trying to find its equation as associated with a coordinate frame attached to the plane of intersection. This new frame has its z axis aligned with the plane normal, but I can choose the origin and the orientation of x axis on the plane to make the equation simpler.
Example scenario
shows a parabloid with the equation:
$$ -0.05x^2-1.1y^2+0.5xy+0.5x+5=z$$ the plane is
$$0.485x-0.485y+0.728z-0.97 = 0$$
Th equation of the resultant parabola(cyan) on the x-y plane is
$$-0.036x^2+0.36xy-0.8y^2+0.85x-0.48y+2.67=0$$
I am trying to find the equation of the white parabola. I have tried projecting 4 points onto the intersection plane and fitting a parabola to it. Those two possible solutions for that are shown in green and red. As can be seen, the red one, which is the closest, doesn't quite match up to the white. I can of course pick a lot more points and try a least square solution, but I was wondering if there is a smarter way to do this.

First, define $\mathbf{r} = [x, y, z]^T $ as the position vector.  Then the equation
$ z= A x^2 + B x y + C y^2 + D x + E y + F $
can be written as
$ \mathbf{r}^T \mathbf{Q r} + \mathbf{b}^T \mathbf{r} + c = 0 \hspace{12pt}(1)$
where
$\mathbf{Q} = \begin{bmatrix} A && \frac{1}{2} B && 0 \\ \frac{1}{2} B && C && 0 \\ 0 && 0 && 0 \end{bmatrix}$
$\mathbf{b} = \begin{bmatrix} D , E , -1 \end{bmatrix}^T $
$c = F $
And the intersecting plane can be written as $ \mathbf{n}^T (\mathbf{r} - \mathbf{r}_0) = 0 $ where
$\mathbf{n} = \begin{bmatrix} a, b, c \end{bmatrix}^T$, and $\mathbf{r_0}$ is any point on the plane, satisfying $\mathbf{n}^T \mathbf{r_0} = - d $
Next, write the vector equation of the plane,
$ \mathbf{r} = \mathbf{r_0} + u_1 \mathbf{v_1} + u_2 \mathbf{v_2}\hspace{12pt} (2) $
where $\mathbf{v_1, v_2}$ are two unit mutually orthogonal vectors are orthogonal to vector $\mathbf{n}$.
Equation (2) can be written compactly as
$ \mathbf{r} = \mathbf{r_0} + \mathbf{V u} \hspace{12pt} (3) $
where $\mathbf{V} = [ \mathbf{v_1}, \mathbf{v_2} ] $ and $ \mathbf{u} = [u_1, u_2]^T$
Now substitute (3) into (1), to obtain,
$ ( \mathbf{r_0} + \mathbf{V u} )^T \mathbf{Q}(\mathbf{r_0} + \mathbf{V u}) + \mathbf{b}^T (\mathbf{r_0} + \mathbf{V u}) + c = 0 $
Expanding,
$ \mathbf{u}^T \mathbf{V}^T \mathbf{Q} \mathbf{V u} + \mathbf{u}^T (2 \mathbf{V}^T \mathbf{Q r_0} + \mathbf{V}^T \mathbf{b} ) + \mathbf{r_0}^T \mathbf{Q r_0} + \mathbf{b}^T\mathbf{r_0}+ c = 0 \hspace{12pt} (4)$
This is the equation of the intersection between the quadric and the plane.
It specifies the relation between the coordinates $u_1$ and $u_2$ in the $\mathbf{r_0 ,V}$ coordinate frame.
The next step is to diagonalize $\mathbf{V}^T \mathbf{Q V} $ so that,
$\mathbf{V}^T \mathbf{Q V} = \mathbf{ R D R}^T $
where $R$ is a $2 \times 2$ rotation matrix and $D $ is a $2 \times 2 $ diagonal matrix.
The above intersection will be a parabola if and only if $\mathbf{D}$ has exactly one diagonal entry equal to $0$.  And we can assume that this zero entry is $D_{22}$.
Define the vector $\mathbf{w} = \mathbf{R}^T \mathbf{u} $, then equation (4) in terms of $w$ becomes
$ \mathbf{w}^T D \mathbf{w} + \mathbf{w}^T \mathbf{g} + h = 0 \hspace{12pt} (5)$
where $ \mathbf{g} = \mathbf{R}^T ( 2 \mathbf{V}^T \mathbf{Q r_0} + \mathbf{V}^T \mathbf{b} ) $ and $ h = \mathbf{r_0}^T \mathbf{Q r_0} +\mathbf{b}^T \mathbf{r_0}+ c $
Remember that $D_{22} = 0 $, and let $D_{11} = a , \mathbf{g} = [g_1, g_2]^T$ and $\mathbf{w} = [x, y]^T $ (not to be confused with the $x$ and $y$ in the world coordinate frame $Oxyz$ ), then equation (5) reads,
$ a x^2 + g_1 x + g_2 y + h = 0 $
For this to be a parabola, we can assume that $g_2 \ne 0$ and $a \ne 0$, then by completing the square, we have
$ a (x + \dfrac{g_1}{2 a} )^2 + g_2 y + h - \dfrac{g_1^2}{4 a} = 0 $
so that,
$ y = -\dfrac{a}{g_2} (x + \dfrac{g_1}{2a} )^2 + \dfrac{1}{g_2} ( \dfrac{g_1^2}{4a} - h ) $
And this specifies the $y$ coordinate as a function of $x$ coordinate in the $\mathbf{r_0},\mathbf{W}$  coordinate frame.
To summarize, at the end we will have two axes given by the columns of the matrix $\mathbf{W} =\mathbf{VR}$. Let these two columns be $\mathbf{w_1} $ and $\mathbf{w_2} $, then the parabola of intersection is given explicitly (and parametrically) by
$\mathbf{r} =\mathbf{r_0} + x \mathbf{w_1} + y \mathbf{w_2} =  \mathbf{r_1}  + t \mathbf{w_1} + \alpha t^2 \mathbf{w_2} \hspace{12pt} (6) $
where the $\alpha = -\dfrac{a}{g_2}$ and $\mathbf{r_1} $ is the vertex of the parabola, given by,
$ \mathbf{r_1} = \mathbf{r_0} + x_0 \mathbf{w_1} + y_0 \mathbf{w_2} \hspace{12pt} (7)$
with $x_0 = - \dfrac{g_1}{2a}, y_0 = \dfrac{1}{g_2} ( \dfrac{g_1^2}{4a} - h ) $
