Solve for the center $O$ of the ellipse:
We know that $O$ is on the circle with $P_1P_2$ as diameter and can verify whether or not $P_3$ is on the corresponding ellipse.
We have:
$$\tag1\langle P_1-O,P_2-O\rangle=0$$
$$\tag2\frac{\langle P_3-O,P_1-O\rangle^2}{\langle P_1-O,P_1-O\rangle^2}+\frac{\langle P_3-O,P_1-O\rangle^2}{\langle P_2-O,P_2-O\rangle^2}=1$$
I am afraid that equation $(2)$ is of two high a degree to allow unique solutions (I bet there are at least four in the general case).
Why not simply use an approximation, after all you will use the ellipse itself as an approximation only? If $M$ is the middle between $P_1$ and $P_3$, assume that the tangent at $P_2$ should be orthogonal to $MP_2$.
If you are looking for the length of the minor and major axis you can calculate $ r_{min} $ and $ r_{max} $ (see formulae below).
If you are trying to determine the bounding box, you can calculate the left-most, right-most, top-most and bottom-most points.
As far as the angle of rotation is concerned, I use the algorithm and formulae below.
Properties of an ellipse from equation for conic sections in general quadratic form
Given the equation for conic sections in general quadratic form: $ a x^2 + b x y + c y^2 + d x + e y + f = 0 $.
The equation represents an ellipse if $ b^2 - 4 a c < 0 $ , or similarly, $ 4 a c - b^2 > 0 $
The coefficient normalizing factor is given by:
$ q = 64 {{f (4 a c - b^2) - a e^2 + b d e - c d^2} \over {(4ac - b^2)^2}} $
The distance between center and focal point (either of the two) is given by:
$ s = {1 \over 4} \sqrt { |q| \sqrt { b^2 + (a - c)^2 }} $
The semi-major axis length is given by:
$ r_\max = {1 \over 8} \sqrt { 2 |q| {\sqrt{b^2 + (a - c)^2} - 2 q (a + c) }} $
The semi-minor axis length is given by:
$ r_\min = \sqrt {{r_\max}^2 - s^2} $
The center of the ellipse is given by:
$ x_\Delta = { b e - 2 c d \over 4 a c - b^2} $
$ y_\Delta = { b d - 2 a e \over 4 a c - b^2} $
The top-most point on the ellipse is given by:
$ y_T = y_\Delta + {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_T = {{-b y_T - d} \over {2 a}} $
The bottom-most point on the ellipse is given by:
$ y_B = y_\Delta - {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_B = {{-b y_B - d} \over {2 a}} $
The left-most point on the ellipse is given by:
$ x_L = x_\Delta - {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_L = {{-b x_L - e} \over {2 c}} $
The right-most point on the ellipse is given by:
$ x_R = x_\Delta + {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_R = {{-b x_R - e} \over {2 c}} $
The angle between x-axis and major axis is given by:
if $ (q a - q c = 0) $ and $ (q b = 0) $ then $ \theta = 0 $
if $ (q a - q c = 0) $ and $ (q b > 0) $ then $ \theta = {1 \over 4} \pi $
if $ (q a - q c = 0) $ and $ (q b < 0) $ then $ \theta = {3 \over 4} \pi $
if $ (q a - q c > 0) $ and $ (q b >= 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} $
if $ (q a - q c > 0) $ and $ (q b < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {\pi} $
if $ (q a - q c < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {1 \over 2}{\pi} $
Best Answer
As far as your diagram looks, you might try some ombination of parabolas to fit your function..
Your lower part (both sides) look like more or less from a same parabola, andthe upper part from a different one. So you might make a function like $f(x)$=Parabola A when y coordinates below a certain point and Parabola B otherwise.
Now what these two (or more) parabolas are, you yourself have to find out , by brute force or something..