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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]}$