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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).

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Ellipses can be "tilted." – André Nicolas Jul 12 '12 at 0:21
By parallel do you mean tangent? Because "parallel" is not quite clear. – Patrick Da Silva Jul 12 '12 at 0:26
@LuanNico: Sort of. Certainly you need only four points to determine an ellipse if it is known that one of its axes is parallel to the $y$-axis. But there is no such thing as the "fifth point." All the points are involved in finding the ellipse. You don't use $4$ points to find a parallel ellipse, and "the fifth" to tilt it. – André Nicolas Jul 12 '12 at 0:34
The general equation for a possibly-rotated conic is $A x^2 + 2 F x y + B y^2 + 2 E x + 2 D y + C = 0$. (The "2"s are for symbolic convenience.) That's six parameters, $A$ through $F$, but you can divide-through by any non-zero one to leave just five. Hence, five points are required to determine such a curve. The conic has axes parallel to the coordinate axes if and only if $F=0$, so that the number of points needed reduces to four. – Blue Jul 12 '12 at 0:39
@Luan Nico: Knowing it is vertical is of no help. Knowing the ratio of the major axis to the minor axis would cut down the points to $3$. – André Nicolas Jul 12 '12 at 0:45
up vote 4 down vote accepted

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.

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@LuanNico: For calculation, it is best not to work with the parameters $x_0$, $y_0$, and $a^2$, $b^2$. The problem is that one gets non-linear equations. It is easier to work with the form $px^2+qy^2+rx+sy+t=0$. Substitute your five points. You get $5$ linear equations in $5$ unknowns, unpleasant but straightforward. – André Nicolas Jul 12 '12 at 1:14

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.

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