So it takes 3 distinct points in the plane, that are not collinear, to define a unique circle that passes through the points.

So what about ellipses?

Arguing naively in terms of degrees of freedom doesn't seem to help too much, because for a circle we have 3 degrees of freedom (2-d center coordinates and 1-d radius), and yet it takes 3 2-d points (6 degrees of freedom), not 2 2-d points (4 degrees of freedom), to define a unique circle that passes through the points. I haven't studied conic sections much so I apologize if this question is trivial, but if we have 4 2-d points in convex position (i.e. each point is a vertex of the convex hull) does this define a unique ellipse that passes through the 4 2-d points? Or do we sometimes need 5 or more 2-d points in convex position?

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    $\begingroup$ The equation for an ellipse in general position is $Ax^2+By^2+Cxy+Dx+EY+F=0$. Of course multiplying by a single constant doesn't change the ellipse, so that's really $5$ variables. Hence you need $5$ values (if nothing else is known about the ellipse). $\endgroup$ – lulu Aug 4 '15 at 20:01
  • $\begingroup$ For any given triangle, there are an infinite number of ellipses which circumscribe it, therefore a lower bound on the number of points required to specify an ellipse on its edge is $4$... $\endgroup$ – abiessu Aug 4 '15 at 20:01
  • $\begingroup$ @lulu: isn't there some Viete trickery available to reduce that equation somewhat? $\endgroup$ – abiessu Aug 4 '15 at 20:02
  • $\begingroup$ @ablessu See the picture posted below. $\endgroup$ – lulu Aug 4 '15 at 20:03
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    $\begingroup$ Have a look at en.m.wikipedia.org/wiki/Five_points_determine_a_conic $\endgroup$ – David Quinn Aug 4 '15 at 20:05

Not quite. Think of the vertices of a square; a tall and narrow ellipse passes through them, but so does a short and wide ellipse. So you don't get uniqueness.

When you (naturally) ask about 5 points, it turns out that there is a unique conic containing any five points...but it's not necessarily an ellipse, even if the points form a nice convex set. Why? Think of 5 points on a parabola. The unique conic that fits these is...that parabola! So there's no ellipse that passes through them.

The projective space of conics is a fascinating introduction to algebraic geometry. The book on Projective Geometry by Pierre Samuel is a pretty nice intro if your other math skills are rock solid, but "other" includes abstract algebra in this case, so it'll be a while before you're ready to read it.

Post-comment remarks

Let me just add another remark here, which hints at the projective geometry thing. Suppose that $$ H_1(x, y) = Ax^2 + 2Bxy + Cy^2 + Dx + Ey + F $$ and that the four points $(x_i, y_i)$, $i = 1, 2, 3, 4$, have the property that $$ H_1(x_i, y_i) = 0 (i = 1, 2, 3, 4). $$ Then the equation $$ H_1(x, y) = 0 $$ defines a conic containing the four points.

Now suppose that $H_2$ is another such polynomial (quadratic in $x$ and $y$) and that the equation $H_2(x, y) = 0$ is satisfied by the same four points. (Think of $H_1 = 0$ as defining the red ellipse in @muaddib's answer, and $H_2 = 0$ as defining the blue one.)

Then for any $t$, the polynomial $$ Q_t(x, y) = (1-t) H_1(x, y) + t H_2(x, y) $$ also is zero at the four points. So in the "space of all conics", if two conics $C_1$ and $C_2$ pass through four points, so do all conics on the "line between $C_1$ and $C_2$" (i.e., those like $Q_t$ above).

To close the argument:

In general, if you have four points $P_1, P_2, P_3, P_4$, and they lie on an ellipse, $E$, you can pick a 5th point, say $R_0$, not close to any of the $P_i$, that's also on that ellipse. Now you can move $R_0$ very slightly to get a new point $R_1$, and consider the conic $C$ passing through $$ P_1, P_2, P_3, P_4, R_1. $$ If $R_1$ is close enough to $R_0$, then the conic $C$ is very close to the original ellipse, and must therefore also be an ellipse (a statement that needs proving, by the way -- it's not at all obvious!).

But now you can take the quadratic for $E$, say $H_1$, and the quadratic for $C$, say $H_2$, and form a combination like $(1-t)H_1 + tH_2$, where $t$ is any real number, to get another conic passing through the four points, so there's a whole infinity of conics passing through these points. By the same argument as above -- the "not at all obvious" one -- infinitely many of these conics must also be ellipses.

In short: no, even in the case where the four points are in "convex position" and there's an ellipse through them, there's always another ellipse through them, and indeed, infinitely many others.

  • $\begingroup$ I guess I'll have to perhaps ask another question then...if 5 points do not necessarily define an ellipse, even in convex position, and if 4 points CAN define an ellipse but it may not necessarily be unique, then what is the middle ground? E.g. if we take 4 points in general position and ask for the ellipse passing through them (assuming they are in convex position) does this uniquely define an ellipse? Or are we in some really weird middle ground where you can find a (non-unique) ellipse through 4 points in convex position, but the 5th point is constrained in order to define an ellipse? $\endgroup$ – user2566092 Aug 4 '15 at 21:12
  • $\begingroup$ @user2566092: Bear in mind that the four-vertices-of-a-square example actually allows for infinitely many ellipses and hyperbolas; so, four points, even "in convex position", aren't going to be ellipse-specific. John's suggestion to look into projective space is a good one, but it lacks this kicker: In projective space, there is only one type of conic. (How that conic interacts with the "line at infinity" is what makes the conic seem to have different flavors to us in non-projective space.) [continued] $\endgroup$ – Blue Aug 4 '15 at 21:37
  • $\begingroup$ @Blue Thanks, but this still doesn't answer my question about "general position" 4 points in convex position, about whether they define a unique ellipse if we insist that the conic section is an ellipse. That's really the thrust (after seeing the answers here) that I'm going for. It seems odd that 4 points in general convex position do not define a unique ellipse, and yet a 5th point in convex position, when added, has to be "special" somehow in order to define a unique ellipse rather than some other kind of conic section. $\endgroup$ – user2566092 Aug 4 '15 at 21:43
  • $\begingroup$ [continued] Since the broader context doesn't distinguish ellipses from parabolas from hyperbolas, you'll have to do better than a collection of points "in convex position" to pin-down one apparent type of curve. Specifically, you'll probably need to exploit the non-projective nature of the Euclidean plane in some way; namely, placing some conditions on the metric properties (distances and angles) defined by your set of points. Ultimately, the condition will be equivalent to the "discriminant" condition of the general quadratic. $\endgroup$ – Blue Aug 4 '15 at 21:44
  • $\begingroup$ @user2566092: Well, again, the four points don't just determine ellipses; there are hyperbolas (and potentially parabolas) that go through the same four points. A fifth point thrown into the mix might land on a member of either family of curves, so you should expect to have to impose a "special" condition on that point to ensure that it lands on a member of the ellipse family. $\endgroup$ – Blue Aug 4 '15 at 21:54

This argument is taken from How many points does it take to define...

The picture says it all:

enter image description here

Suppose that we have K points, and we want these points to define only a single ellipse. If the points do define two ellipses, then the points must be found on the circumference of both – so they must be on the intersections between the two ellipses. Two ellipses intersect at most at four points – so if we have only four points, we are not able to resolve the ambiguity between the two. In order for there to be a contradiction, there must be at least five distinct points. Note that this proof doesn’t say that five points are enough; it only says that four are not.


This is an image to accompany a point I was making in a comment:

enter image description here

Four points ---here, the cyan ones at the corners of a square near the center--- don't just determine infinitely-many ellipses; they determine infinitely-many hyperbolas, too. (No parabolas this time.) Moreover, those ellipses and hyperbolas cover disjoint regions of the plane, in such a way that each point on the plane (other than the first four) lies on exactly one curve. This is what it means for a fifth point added to the collection to determine one of those curves uniquely. This is also why there has to be a "special" condition (namely: in which region to place the point) to guarantee that the fifth point determines the specific type of curve (ellipse or hyperbola) you might want.

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    $\begingroup$ Great picture! Note to OP: a projective transformation can move those four cyan points to any other four points of the plane, so this picture is more or less the same for any four points you choose. $\endgroup$ – John Hughes Aug 5 '15 at 0:06

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