# Ellipse issue

• Sep 30th 2008, 04:34 PM
DangerousDave
Ellipse issue
Hi,

I have an unusual problem. I want to know the ratio of the height of an ellipse from either focus compared with the height of the ellipse at the centre, as the distance between the foci tends to infinity. When the distance between the foci tends to zero, the height above (below) the foci tends to the height at the centre because the foci tend to the centre. But I'm wondering if, as the distance tends to infinity, the foci height tends to half the centre height. Any help greatly appreciated - this matter has philosophical import!!!

DD
• Sep 30th 2008, 05:13 PM
icemanfan
After doing a bit of scratch work, I determined that as an ellipse gets flatter, the ratio of the focus height to the height at the center tends to zero. In other words, you can find an ellipse to make this ratio as small as you would like.
• Sep 30th 2008, 06:38 PM
DangerousDave
I'm not entirely sure what you mean when you say the ratio tends zero. I was kind of looking for an x:y type expression of a ratio. Do you mean that the height at the centre is infinitely greater than the height at the foci for an infinitely wide ellipse? I'm not sure that works out. That would make for a sort of galaxy (from the side) shaped ellipse. Actually I think I should amend the question to clear it up: the ratio between focus height and centre height for an ellipse whose "width" is infinitely greater than its "height".

Cheers
• Sep 30th 2008, 07:05 PM
icemanfan
Quote:

Originally Posted by DangerousDave
Do you mean that the height at the centre is infinitely greater than the height at the foci for an infinitely wide ellipse?

Yes, that is exactly what I mean. Consider an ellipse centered at the origin, with foci (-d, 0) and (d, 0), and the sum of the distances to the foci equal to c. Then the equation of the ellipse is $\frac{4x^2}{c^2} + \frac{4y^2}{c^2 - 4d^2} = 1$. For a "flat" ellipse, let c approach 2d.

At x = d:

$\frac{4d^2}{(2d + \delta)^2} + \frac{4y_d^2}{(2d + \delta)^2 - 4d^2} = 1$, which simplifies to

$y_d^2 = \left(1 - \frac{4d^2}{(2d+\delta)^2}\right)\cdot \left(\frac{(2d + \delta)^2 - 4d^2}{4}\right)$

At x = 0:

$\frac{4y_0^2}{(2d + \delta)^2 - 4d^2} = 1$

$y_0^2 = \frac{(2d + \delta)^2 - 4d^2}{4}$

To summarize:

$\frac{y_d^2}{y_0^2} = 1 - \frac{4d^2}{(2d+\delta)^2}$

$\frac{y_d}{y_0} = \sqrt{1 - \frac{4d^2}{(2d+\delta)^2}}$

Depending on our delta, that factor can be as small as we want it to be.
• Oct 5th 2008, 10:12 AM
DangerousDave
Hi - Just got back to this problem. Thanks for your response. It's a long time since I was in maths class so excuse my slowness. I was reading through and one thing that struck me is that if the distance between the foci is c and the foci are at (0, d), (0, -d) then c always = 2d. So i'm not sure about letting c approach 2d.
The other thing i'm currently puzzling over is how you got your equation there. Wikipedia says that the equation of an origin-centred ellipse is:

x^2/a^2 + y^2/b^2 = 1

where a is the semimajor axis and b is the semiminor axis. Gosh they have terms for the things I'm looking for! The ratio between the semiminor axis and the semilatus rectum when d approaches the semimajor axis. And if that ain't a headache I don't know what is. Apparently:

al = b^2

where l is the semilatus rectum. But that doesn't really help because the equation doesn't involve that d value, the distance from the centre to the foci. We want to let d tend to a. I'm going to draw some more pictures and think about things. Cheers
• Oct 5th 2008, 10:37 AM
DangerousDave
In fact I can't find any equation involving the foci at all. I was thinking about using trigonometry because of that 2 pins and a piece of string way to create an ellipse. I had it all drawn out but my trigonometry isn't good enough these days. I hope I just attached a picture. Anyone make anything of this? I want to know the relationship between b and l when w/2 tends to a.
• Oct 5th 2008, 12:10 PM
earboth
Quote:

Originally Posted by DangerousDave
In fact I can't find any equation involving the foci at all. I was thinking about using trigonometry because of that 2 pins and a piece of string way to create an ellipse. I had it all drawn out but my trigonometry isn't good enough these days. I hope I just attached a picture. Anyone make anything of this? I want to know the relationship between b and l when w/2 tends to a.

Consider an ellipse centered at the origin with the major semi-axis a and the minor semi-axis b. Then the foci are at $F_1\left(-\sqrt{a^2-b^2}, 0 \right)$ and $F_2\left(\sqrt{a^2-b^2}, 0 \right)$.

The height above the focus is the y-value if you plug in $x = \sqrt{a^2-b^2}$into the equation of the ellipse. You'll get $y = \frac{b^2}{|a|}$.

So if b stays constant and a is approaching infinity the height above the focus will approach zero.