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

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for $a=b$ i.e. a circle the answer is at most $2$, depends on the distance between the points. –  simco Oct 10 '12 at 12:16

3 Answers 3

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):

$\hskip{2cm}$Ellipse diagram

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.

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Well, I've just found out this post: stackoverflow.com/questions/197649/… . A suggested answer reports that the solution could be found, if it does exist (small $d$ as you explained). In this case, it should be at least two solutions. For infinitely many, I am not sure, according to the post, probably not. –  simco Oct 10 '12 at 12:35
    
The post referenced does assume that the ellipse is not rotated WRT the coordinate system. I didn't intend to impose such conditions. My idea is similar to what's explained in the other answer. –  Lord_Farin Oct 10 '12 at 21:14
    
Thank you, I got it. Could you define what kind of rotation it is such that those certain points still lie on the ellipse? What is the center of this rotation? –  simco Oct 12 '12 at 9:42
    
You may not have understood entirely what I meant - I meant to say that if we insist the ellipse be aligned with the coordinate axes, we get finitely many solutions. My remark referred to the fact that you get more solutions if you don't require this alignment, i.e. rotate the ellipse by a bit (WRT coordinate axes). I don't think that in general the solutions for such a "rotated" ellipse will have the ellipse centered at the same point (thus we may need to move around the ellipse after rotating it to get the desired solution). I'll try to prove it rigorously with analytic methods, later today. –  Lord_Farin Oct 12 '12 at 16:16

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.

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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.

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Whoah, wait -- how do you know that they won't all rotate into the same ellipse? (I know they won't, but there are infinitely many that will.) –  TonyK Oct 14 '12 at 10:22
    
There is an infinite family of ellipses that can be formed by "sliding" some given ellipse along the point P. After rotation to touch Q, members of this family will certainly be different ellipses (since their contacts with P are different). That's what I had in mind, though admittedly not quite what I wrote. –  bubba Oct 14 '12 at 10:38

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