Number of ellipses through two fixed points in 2D space?

Question

How many ellipses with a given size (mean $a$ and $b$ given) one can draw through two fixed points in 2D plane?

Answer 1

Suppose that $a \ne b$, i.e. the ellipse is not a circle, and WLOG put $a > b$.

Then if the distance $d$ between the given points is $2a$, the answer is evidently just $1$; if $d > 2a$, evidently no such ellipse exists.

If the distance is smaller, since the ellipse does not have rotational symmetry, I suspect the answer to be infinitely many (just by "wiggling" a solution around a bit). This is of course not very rigorous, but I suspect it can be made formal.

EDIT: A formal proof of the infinitude of solutions (also yielding some other things):

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The parallelogram in the ellipse indicates that (as ellipses are convex) the maximum length of a line segment along the coordinate axes inside a rotated ellipse must go through the center of the ellipse (for the parallelogram has 180 degree rotational symmetry). Say this maximal length for an ellipse rotated by $\theta$ WRT coordinate axes is $d(\theta)$. I saved myself the effort of proving that any length between $0$ and $d(\theta)$ may be attained, but this is apparent from the diagram.

Now $d(\theta)$ is just twice the "radius" $r(\theta)$ (i.e., the distance from the center to the actual ellipse in the direction $\theta$) which is given by:

$$r(\theta) = \sqrt{(a \cos\theta)^2+(b \sin\theta)^2}$$

I will show that $r(\theta)$ (and so $d(\theta)$) decreases as $\theta$ increases from $0$ to $\pi/2$. Together with the continuity observed earlier this will give the result.

By the montonicity of the square root function it suffices to show that $r(\theta)^2$ decreases; to this end compute the derivative:

$$\frac{d}{d\theta}r(\theta)^2 = -2a^2\sin\theta\cos\theta + 2b^2\sin\theta\cos\theta = \sin(2\theta)(b^2-a^2)$$

where I used $2\sin\theta\cos\theta = \sin(2\theta)$. But $\sin(2\theta)$ is positive for $0<\theta<\pi/2$. Since $a > b$, we conclude that $r(\theta)$ is decreasing.

Now if we are given a distance $d < 2a$ between our points, by the intermediate value theorem we find $0<\theta<\pi/2$ such that $d \le d(\theta)$. We observed from the diagram (essentially a continuity argument) that any length between $0$ and $d(\theta)$ could be attained in an ellipse rotated by $\theta$; in particular, this shows $d$ can be attained.

Since $\theta >0$, this gives us solutions for all $0 \le \theta' \le \theta$, and this last interval has the cardinality of the continuum.

In conclusion, there are infinitely many solutions. I hope the argument is clear and insightful; IMHO the explicit determination of the assertion about continuity would obfuscate the point of the proof. Certainly, it would be very tedious.

Answer 2

There are an infinite number if the distance between the points is less than $2a$. Think of drawing a chord across the ellipse of length of the distance between the points. Each chord corresponds to a different ellipse through the points. There is at least a region around the end of the major axis where this is possible.

Answer 3

Infinitely many, provided the distance between the points is fairly small.

Let's call the points P and Q. Obviously there are infinitely many ellipses of size $a \times b$ that can be drawn through P. Any one of these ellipses can be rotated around P until it touches the other point Q.

Another way to think about it ...

Take some ellipse of size $a \times b$ that touches P and Q. It seems obvious that you can then "slide" this ellipse back and forth, keeping it in contact with P and Q. So, again, an infinite number of solutions. Maybe this doesn't qualify as a mathematical proof, but the physical process is clear enough, so it seems convincing to me.