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I have seen it stated that you need 5 points to define an ellipse, and that an ellipse can be drawn through 5 points as long as any $3$ points aren't on the same line. How can an ellipse be drawn through these $5$ points: or am I missing some other condition on when ellipses can be formed from a set of points $(1,1),(1,-1),(-1,1),(-1,1),(0,1/2)$?

(1,1),(1,-1),(-1,1),(-1,1),(0,1/2)

TShiong
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Jack
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    It defines an ellipse, with imaginary parameters. In other words, it defines a hyperbola. – Trebor Mar 08 '22 at 14:18
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    It defines an ellipse that goes past infinity in one direction and loops over to the other side. In other words, it defines a hyperbola. – Arthur Mar 08 '22 at 14:30

2 Answers2

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If you have 5 points, no three collinear, then there is a unique conic section that goes through them. That includes ellipses, but also opens up the possibilities of parabolas and hyperbolas.

Only hyperbolas can cover a non-convex set of 5 points like yours. In your particular case, assuming the points are exactly where it looks like they are, that hyperbola is given by $$ 4y^2-3 x^2=1 $$

Arthur
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Five points define at most one ellipse.

But it's not true that any five points define an ellipse.

So if you have five points and an ellipse can be drawn through them, then this ellipse is unique. That is the thing here.

More info can be found here:

How many points are needed to uniquely define an ellipse?

peter.petrov
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