April 26th, 2016, 04:38 AM  #1 
Member Joined: Jun 2012 Posts: 73 Thanks: 2  Find the length of parts of an ellipse
Hi, I have an ellipse, where radius 1 and radius 2 are known variables. I know an angle alpha, which intersects the ellipse at point T. From that point T, I would like to draw a tangent line, spanning by known distance L from both sides of the point T. Ends of this tangent line will cut the ellipse at points S1 and S2. Is it possible to calculate the lengths of S1_T and S2_T ellipse parts (colored in green on photo attached below)? So what I have: radius 1 radius 2 alpha L What I would like to calculate is the length of ellipse parts: S1_T S2_T I would appreciate any kind of help. Thank you. 
April 26th, 2016, 11:49 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,624 Thanks: 2077 
Will approximate values suffice? They can be obtained by use of calculus.

April 27th, 2016, 12:59 AM  #3 
Member Joined: Jun 2012 Posts: 73 Thanks: 2  
April 27th, 2016, 07:26 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,624 Thanks: 2077 
It's not possible except in very special cases, such as when the ellipse is a circle.

April 27th, 2016, 10:25 AM  #5 
Member Joined: Jun 2012 Posts: 73 Thanks: 2 
Thank you skipjack. I am interested in ellipse example where radius 1 and radius 2 are not equal. Can you show me the calculation of those approximate values, in case radius 1 and radius 2 are not equal? 
April 30th, 2016, 02:26 PM  #6  
Member Joined: Jun 2012 Posts: 73 Thanks: 2 
@ skipjack: Did I misunderstand you when you said: Quote:
Is this correct? Last edited by stgeorge; April 30th, 2016 at 02:44 PM.  
April 30th, 2016, 03:38 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,624 Thanks: 2077 
No, an ellipse. You will need an equation for the ellipse.

May 1st, 2016, 08:40 AM  #8 
Member Joined: Jun 2012 Posts: 73 Thanks: 2 
Hi @skipjack. Can you present that approximate solution please? Thank you. 
May 5th, 2016, 01:46 AM  #9 
Member Joined: Jun 2012 Posts: 73 Thanks: 2 
Hi @skipjack, Did I do something wrong? Let's assume that radius 1 (semimajor axis) length is 8, and that radius 2 (semiminor axis) length is 5. I assume that an ellipse equation would like like: x^2/8^2 + y^2/5^2 = 1 Can you show me how would those those approximate values obtained through calculus look like? Thank you in advance. Last edited by skipjack; May 5th, 2016 at 04:31 AM. 
May 8th, 2016, 04:52 PM  #10 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications 
Hi stgeorge, I would like to take the liberty of answering your question. I limited myself to quadrant 1 (I think) so be aware of that. Let's work an example with the numbers for $radius_1$ and $radius_2$ that you chose and arbitrary (I chose) $alpha$ and $L$ : $\displaystyle \large { \begin{align*} radius_1 &=8 \\ radius_2 &=5 \\ alpha &= \frac{\pi}{5} \\ L &=2 \end{align*} }$ It is easiest (in my opinion) to work in parametric equations with the ellipse defined as: $\displaystyle \large{ x=radius_1 \cdot \cos(t) \\ y=radius_2 \cdot \sin(t)}$ as referenced here in equations (13) and (14). Note that the parameter $t$ is not the same as the angle from the origin to a point on the ellipse. However, we can calculate the relationship between t and an arbitrary alpha, $aplha_{arb}$ as follows: $\displaystyle \large{ \begin{align*} \tan(alpha_{arb})&=\frac{y}{x} \\ \tan(alpha_{arb})&=\frac{radius_2 \cdot \sin(t)}{radius_1 \cdot \cos(t)} =\frac{radius_2}{radius_1} \cdot \tan(t) \\ \tan(t)&=\frac{radius_1}{radius_2} \cdot \tan(alpha_{arb}) \\ t&=\arctan\left (\frac{radius_1}{radius_2} \cdot \tan(alpha_{arb}) \right) \end{align*}}$ So at point $T$ : $\displaystyle \large{ \begin{align*} t_{alpha}&=\arctan \left( \frac{8}{5} \cdot \tan \left( \frac{\pi}{5} \right) \right) \approx 0.86038812 \quad [1] \\ x_T&=8 \cos(t_{alpha}) \approx 5.217146282 \\ y_T&=5 \sin(t_{alpha}) \approx 3.790478649 \end{align*}}$ Now for a little calculus. The derivative of the curve, which gives the slope of the line tangent to the curve is: $\displaystyle \large{ \begin{align*} \frac{dy}{dt}&=5 \cos(t) \\ \frac{dx}{dt}&=8 \sin(t) \\ \frac{dy}{dx}&=\frac{dy/dt}{dx/dt}=\frac{5 \cos(t)}{8 \sin(t)}=\frac{5}{8}\ \cot(t) \end{align*}}$ So: $\displaystyle \large{ \begin{align*} T_{slope}&=\frac{5}{8} \cot(0.86038812) \approx 0.537649188 \\ x1_{L_{S1}}&=5.2171462822\cos(\arctan(0.537649188 )) \approx 3.455607093 \\ y1_{L_{S1}}&=3.7904786492\sin(\arctan(0.537649188 )) \approx 4.737568763 \end{align*}}$ Note that we don't have to calculate the actual coordinates of S1. But we need $t_{S1}$ . Fortunately we have the tangent of the angle wrt the origin as $y1_{L_{S1}}/x1_{L_{S1}}$ . So we have: $\displaystyle \large{ t_{S1}=\arctan \left( \frac{8}{5} \cdot \frac{4.737568763}{3.455607093} \right) \approx 1.143064833 \quad [2] }$ Similarly for $S2$ : $\displaystyle \large{ \begin{align*} x2_{L_{S2}}&=5.217146282+2\cos(\arctan(0.537649188 )) \approx 6.978685471 \\ y2_{L_{S2}}&=3.790478649+2\sin(\arctan(0.537649188 )) \approx 2.843388535 \\ t_{S2}&=\arctan \left( \frac{8}{5} \cdot \frac{2.843388535}{6.978685471} \right) \approx 0.577711407 \quad [3] \end{align*}}$  Now for more calculus to find the arc lengths. If, for ease of notation, we let: $\displaystyle \large{ a=radius_1 \\ b=radius_2}$ The differential arc length, $\Delta s$, and arc length $s$, are given by: $\displaystyle \large{ \begin{align*} \Delta s &= \sqrt{\Delta x^2+\Delta y^2} \\ s &= \int_{t1}^{t2} \sqrt{a^2 \sin^2t+b^2 \cos^2t} \ \ dt \\ s &= \int_{t1}^{t2} \sqrt{b^2 \sin^2t+b^2 \cos^2t + \left(a^2b^2 \right)\sin^2t} \ \ dt \\ s &= \int_{t1}^{t2} \sqrt{b^2 + \left(a^2b^2 \right)\sin^2t} \ \ dt \\ s &= \int_{t1}^{t2} \sqrt{b^2 + b^2 \frac{\left(a^2b^2 \right)}{b^2}\sin^2t} \ \ dt \\ s &= b \ \int_{t1}^{t2} \sqrt{1 + \frac{\left(a^2b^2 \right)}{b^2}\sin^2t} \ \ dt \\ s &= b \ \int_{t1}^{t2} \sqrt{1  \frac{\left(b^2a^2 \right)}{b^2}\sin^2t} \ \ dt \\ s &= b \ \int_{t1}^{t2} \sqrt{1  \left(1\frac{a^2}{b^2} \right) \sin^2t} \ \ dt \\ \end{align*}}$ This is an elliptic integral and fortunately, WA can provide the answers. For S1_T, using values [2] and [1], look here For S2_T, using values [3] and [1], look here. Note that in these equations, I used $1\frac{a^2}{b^2}$ in the EllipticE function while equation (57) in the WMW reference given above, erroneously, I think, uses the square root of this value. I tried the square root and came up with unreasonable values. I am quite sure that the results using the non square root value are correct as I used two methods of integration to verify the answers. It is easy to make mistakes so I wrote a program to calculate these values and do the integration. I will post the code shortly. It is in Ruby but one should be able to convert it to other languages. Last edited by greg1313; May 10th, 2016 at 05:55 PM. 

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ellipse, find, length, parts 
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