How many points are necessary to find a parallel ellipse, and how to do it? So, I understand that to find an ellipse for sure you need at least five points. Why? The ellipse equation has only four variables ($x_0, y_0, a,\text{ and }b$). That's not actually my true question, just a curiosity. My question is, what if I know that the ellipse is parallel to the y-axis---how many points then do I need to find it, and how to do it? I have three, but I can get to four if necessary. The problem is I just can't see, on the ellipse formula, how to indicate that it is parallel, and I keep with 3 variables and 4 equations, because I can't 'insert' this information! By using 'find and ellipse' I mean finding it's center and axes length (that is, $x_0, y_0, a\text{ and }b$). By points I mean points that the ellipse contains (that is, x and y in the formula).
 A: I assume that you mean major or minor axis of ellipse is parallel to y-axis.
The equation of an ellipse whose major and minor axes coincide with the Cartesian axes is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
For your case you only need to shift the origin to some other point say $(x_0,y_0)$ and equation of ellipse transforms as $$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$$
Evidently you need four variables to solve it for $x_0,y_0,a,b$.
On a side note for a general ellipse you need to rotate the co-ordinate axes and also shift the origin to some other point.
suppose you rotate by angle $\theta$ anticlockwise and shift origin to $(x_0,y_0)$ then 
$x\mapsto((x-x_0)\cos\theta+(y-y_0)\sin\theta)$ and $y\mapsto(-(x-x_0)\sin\theta+(y-y_0)\cos\theta)$ then equation of ellipse transforms as 
$$\frac{((x-x_0)\cos\theta+(y-y_0)\sin\theta)^2}{a^2}+\frac{(-(x-x_0)\sin\theta+(y-y_0)\cos\theta)^2}{b^2}=1$$
Above has five variable so you would need five points.
A: There are four constants in a "parallel" ellipse so you need four constants to write such an ellipse.
For the  "parallel" or better "axes parallel" ellipse (  $x,y$ axes are parallel to directions of lines $a,b$ ) the coefficient for $xy $ term vanishes. Comparing term by term in the equation establishes equivalence of two ways of representation:
$$ \left(\frac{x-h}{a}\right)^2+\left(\frac{y-k}{b}\right)^2-1 =0  $$
$$ A x^2 + B y^2 + 2 E x + 2 F y -C = 0 $$
By comparing coefficients for the four corresponding variables, the constants equivalence is:
$$ A = \dfrac{1}{a^2}\, ; B= \dfrac{1}{b^2} \,; E = \dfrac{-h}{a^2} \,; F = \dfrac{-k}{b^2}\,; \, C=1-\dfrac{h^2}{a^2}  -\dfrac{k^2}{b^2} \,;$$
which appear to be five. However you can divide the coefficients in the last equation by $C$ for full agreement.
For the tilted ellipse you need 5 constants and 5 points.
