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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."2012-07-12
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    By parallel do you mean tangent? Because "parallel" is not quite clear.2012-07-12
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    @André Nicolas If I got it right, you are saying that the ellipse formula can just be used to find parallel ellipses, and that the 5th point is used to determinate it's rotation?2012-07-12
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    @Patrick Da Silva By parallel I mean that it's axis are parallel to the coordinate system's axis, so, that is not rotated.2012-07-12
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    @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.2012-07-12
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    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.2012-07-12
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    It's really quite simple. The equation of an ellipse with axes parallel to the coordinate axes looks like the following in its most general form $$\left(\frac{x-h}{a}\right)^2+\left(\frac{y-k}{b}\right)^2=1$$ where $a$ and $b$ are the major and minor axes, and $(h,k)$ is the center of the ellipse. That's four parameters total.2012-07-12
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    OK, I think I understood (the ellipse formula I knew was different). Thanks! And one more thing, can I reduce it to three points if I know that the ellipse is 'vertical', that is, the y-axis is larger than the x one?2012-07-12
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    @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 $3$.2012-07-12
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    OK, thank you all very much! I'm sure I'll be able to solve my problem the best way now. By the way, if any of you want, you can really answer the question (instead of posting comments) so I can upvote and choose your answer!2012-07-12
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    sabertooth's answer is a good summary of all the things we said; you can upvote/accept his/hers?2012-07-12
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    @ Luan NicoYou mean a parallel ellipse has its axes parallel to coordinate axes?2016-02-09
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    @Narasimham yep, that's what I meant. Sabertooth's answer pretty much nails it, take a look ;)2016-02-09

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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.2012-07-12
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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.