# What is a simple way to locate the foci of an ellipse?

I have DIY project going, and ran into the same issue as posted in the question here: stove jack for a tent

The answer was helpful, but somewhere along the line I got lost in the math... (Perhaps because the example was over simplified, using a 45 triangle with units of 1, and the distance between foci equaled the minor axis?)

Because my roof pitch isn't 45 degrees, I wanted to work the problem on a 30-60-90 triangle to make sure I understood the concept before I unfold my tent and take measurements of the actual slope. I plan on using a 4" diameter stove pipe, and created the diagram below. (apologies for the image quality) I realize that I could make a template using a French curve like I did for my diagram above and be "close enough", but I would like to try the string method as described in the answer on the other question I linked to.

Is there a relatively easy way to calculate distance X given the angles, and/or dimensions for someone like me who is not a math whiz?

I am accepting Parcly Taxel’s answer because it was concise, first, correct, and it pointed me in the right direction even if it didn’t state the obvious.

I understand that equations are the language of mathematics, and I didn’t ask for a plain English answer, but as I alluded to in a comment – for many of us knowledge of anything beyond the basic arithmetic we use regularly may have been hard won, and could have atrophied in the intervening decades. Even equations that are very simple for the left brained intelligentsia frequenting this venue may create anxiety for us simpletons cutting holes in our tents who need to review whether to use COS or TAN to calculate the opposite side!

I don’t want this to morph into a critique about how I was taught math, but now that I “get it” I wanted to offer an example below of how I might answer the question for an admitted non-math person. FWIW, and I welcome any feedback in the comments…

• I think you probably know how to accept answers, no? Oct 26 '20 at 15:15
• Yes, I do. I have my reasons for holding off right at this moment though... Oct 26 '20 at 15:18
• @Parcly Taxel, I have accepted your answer. Sorry to sound mysterious, but I have seen feedback on other exchanges that it is considered impolite to accept an answer too quickly because others may have a better answer but choose not to post it as a result. Again, thanks! (And let me know what you think about my addendum and self answer...) Oct 26 '20 at 18:03

Say the pipe has radius $$r$$ and the roof has angle $$\theta$$ to the horizontal. The ellipse's semi-axes will be $$b=r$$ and $$a=r\sec\theta$$. Now, checking Wikipedia we find

Linear eccentricity
This is the distance from the center to a focus: $$c=\sqrt{a^2-b^2}$$.

So the distance between foci in our case will be $$2\sqrt{r^2(\sec^2\theta-1)}=2r\tan\theta$$.

• So by this formula, the distance between foci will always equal one of the legs of the right triangle. (in this case, the side opposite the 30 degree corner) Is this true and consistent for all angles? Because it contradicts the other answer... Oct 26 '20 at 14:53
• @MichaelHall There's no contradiction if my $r$ is John Hughes's $r_1$ (true) and my $\theta$ is John's $A$ (true). And this is true for all angles and radii. Oct 26 '20 at 14:58
• There was, but he changed his answer... Thanks BTW! Oct 26 '20 at 15:02

Using standard symbols of an ellipse:

For A=$$30^0$$

$$a= b \sec A\quad = b \cdot \dfrac{2}{\sqrt3}; \quad c^2=a^2-b^2 =\dfrac{b^2}{3}$$

Take square root and double it for inter-focal distance

$$c=\dfrac{b}{\sqrt3},\quad 2c =\dfrac{2b}{\sqrt3} \approx 2.31\text{ for b=}2^{''.}$$

4-inch pipe means that the narrow axis $$r_1$$ of the ellipse in your picture is $$2$$. In general, the long axis (in terms of the angle $$A$$ at the lower right, so $$A = 30$$ degrees in your picture) is going to be \begin{align} r_2 &= \frac{r_1}{\cos A} \end{align} where the cosine must be computed for degrees on your calculator.

The distance from the center fo the ellipse to one of the foci is $$d = \sqrt{r_2^2 - r_1^2}$$ so the number you need is twice that: \begin{align} D &= 2 \sqrt{r_2^2 - r_1^2}\\ &= 2 \sqrt{\left(\frac{r_1}{\cos A}\right)^2 - r_1^2}\\ &= 2r_1 \sqrt{\left(\frac{1}{\cos A}\right)^2 - 1}\\ &= 2r_1 \sqrt{\sec^2 A - 1}\\ &= 2r_1 \sqrt{\tan^2 A}\\ &= 2r_1 \tan A \end{align} In our case, $$\tan A = \tan 30^\circ \approx .577$$, and $$r_1 = 2$$, so we get $$D \approx 4 \cdot 0.577 = 2.309.$$

• OK, I see you edited your answer so they now both match. FYI, this is probably child's play for you, but the formulas above are giving me flashbacks to my math class struggles decades ago! Thanks for the detailed answer, but I like the simplicity of the other one better. Oct 26 '20 at 15:01
• Totally fine. It is pretty easy for someone who's been doing mathematics for 50 years, but it's also easy to screw up (as my initial answer showed...I'm still not sure where I lost a factor of $2$ offhand!). I agree the other is simpler; mine gives more details of the steps of the algebra. I'm not at all insulted by your choosing @ParclyTaxel's answer, which was the first to be correct, after all. Oct 26 '20 at 15:39