Determining the cartesian equation of an ellipse ( be it neither in standard position nor orientation) given center, one vertex and one semi -axis Desmos construction, with sliders waiting for being launched  : https://www.desmos.com/calculator/vuou1gnese


Question : I obtained the final formula while thinking of center $C$
and vertex $V$ as points lying in the first quadrant, with $V$ located
to the right  of $C$  Is the formula general in spite of this? - I
assumed that the " shifting terms" $h$ and $k$ operate , so to say, in
the rotated system, so that,  intead of substituting $X_C$ for $h$ and
$Y_C$ for $k$ , I used the coordinates of center $C$ in the rotated
system. Is this assumption correct? Finally, can you think of other forms of this  equation?

The givens are :

*

*$\text {Center} = C = (X_C, Y_C) $


*$ \text {Vertex} = V = ( X_V,Y_V) = (X_C+p, X_C+q) , \text {with }p, q$
$\in \mathbb R$


*$ m {\in\space \mathbb R^+}  = \text {length of one semi  axis ( perpendicular to the semi-axis [CV] }$
From what is given can be immediately deduced :

*

*$ M = \text {length of the semi axis [CV]}= \sqrt  {p^2 + q^2}$


*$R = \text {inclination of the straight line (CV) on which lies the semi axis [CV] }= \arctan {\frac qp}$


*the equations of $X'$ and $Y'$, the rotated coordinate system in which lies the parabola , namely
$\space \space \space \space  X' :   y = \tan(R)x $
$\space \space \space \space  Y' :   y = (-1 / \tan(R))x $
Since the ellipse lies in a rotated coordinate system, and since its center is not necessarily at the origin its equation will be of the form :
$$\frac { (X(x,y) -h )^2 }  { M^2 } + \frac { (Y(x,y)-k )^2  }  { m^2 } =1$$
with

*

*$X(x,y) = x \cos(R) + y \sin (R) $


*$Y(x,y) = y \cos(R) - y \sin (R) $


*$ h$ and $k$ : the coordinates of center $C$ in the rotated coordinate system.
Determining the values of $h$ and $k$

*

*$h =  X_C \cos (R) + Y_C \sin (R) $


*$k = Y_C \sin (R) - X_C \cos(R)$
Hence the formula for the ellipse of center $C= (X_C , Y_C) $, vertex $V$ and a semi-axis of length $m$ is :
$$\frac { (X(x,y) - (X_C \cos (R) + Y_C \sin (R)) )^2 }  { M^2 } + \frac {( Y(x,y)-(Y_C \sin (R) - X_C \cos(R))  ) ^2 }  { m^2 } =1$$
with, as said above $ M= \text {length of the semi- axis [CV]}$
 A: Too big for a comment, but using the composition of arctan and sine/cosine actually simplifies things a lot more than I expected, because $M$ reappears. The identities are:
$$\sin \arctan \frac q p = \frac{\frac q p}{\sqrt{1+\left(\frac q p\right)^2}} = \frac p {\sqrt{p^2+q^2}} = \frac p M \\
\cos \arctan \frac q p = \frac{1}{\sqrt{1+\left(\frac q p\right)^2}} = \frac q {\sqrt{p^2+q^2}} = \frac q M$$
This makes your final equation at the bottom:
$$\frac{1}{M^2} \left( \left( \frac{p}{M}x + \frac{q}{M}y \right) - \left( \frac{p}{M}X_C + \frac{q}{M}Y_C \right) \right)^2  +  \frac{1}{m^2} \left( \left( \frac{q}{M}y - \frac{p}{M}x \right) - \left( \frac{q}{M}X_C - \frac{p}{M}Y_C \right) \right)^2  = 1 \\
\frac{1}{M^4} \left( px + qy - pX_C - qY_C \right)^2  +  \frac{1}{M^2m^2} \left(  py -qx  -  pY_C + qX_C \right)^2  = 1 \\
\frac{1}{M^2} \left( px + qy - pX_C - qY_C \right)^2  +  \frac{1}{m^2} \left(  py -qx  -  pY_C + qX_C \right)^2  = M^2$$
I fear that further "simplification" just ends up with huge constants just to put it into $Ax^2+By^2+Cxy+Dx+Ey+F$ form.
A: Rotate then Shift.
Starting from the algebraic equation of an ellipse in standard format
$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $
If you define the position vector $\mathbf{p} = [x, y]^T $ , then the above equation can be written concisely as follows
$ \mathbf{p}^T \ D \ \mathbf{p} = 1\hspace{50pt}(*)$
where
$ D = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix} $
Now in the end, you want to find the algebraic equation of the same ellipse but rotated by some angle $\theta$ (with respect to the original orientation) and shifted by a certain displacement vector $(Xc, Yc)$
So, first rotate the ellipse about the origin, then shift the resulting the ellipse by the vector $(Xc , Yc)$
The image of a point $\mathbf{p} = (x,y)$ on the original ellipse under a rotation by an angle $\theta$ is the point
$\mathbf{p'} = R \ \mathbf{p} $
where $R$ is the two-dimensional rotation matrix given by
$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
And the image of $\mathbf{p'} $ after a shift by $C = [X_c, Y_c]^T $ is
$ \mathbf{p''} = \mathbf{p'} + \mathbf{C} = R \ \mathbf{p} + \mathbf{C} $
From this last equation, we deduce that $ \mathbf{p} = R^T (\mathbf{p''} - \mathbf{C} ) $
Substituting this into equation $(*)$, gives us
$ (\mathbf{p''} - \mathbf{C})^T \ R D R^T \ (\mathbf{p''} - \mathbf{C} ) = 1 \hspace{50pt} (**)$
And this is the algebraic equation of the rotated/shifted ellipse.
Explicitly, we have
$ R D R^T = \begin{bmatrix} \dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta && \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta \\ \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta && \dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta \end{bmatrix}$
And $\mathbf{p''} - \mathbf{C} = (x - X_c, y - Y_c) $
Therefore, the algebraic equation is
$\bigg(\dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta \bigg) (x - X_c)^2 +  2 \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta
(x - X_c) (y - Y_c) + \bigg(\dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta\bigg) (y - Y_c)^2 = 1 $
Using the double angle trigonometric formulas, this last equation can be expressed as follows
$(A + B \cos(2 \theta)) (x - X_c)^2 +  2 B \sin(2 \theta) (x - X_c) (y - Y_c) + (A - B \cos(2 \theta)) (y - Y_c)^2 = 1 $
where
$ A = \dfrac{1}{2} \bigg( \dfrac{1}{a^2} + \dfrac{1}{b^2} \bigg) $
$ B = \dfrac{1}{2} \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) $
