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Ellipse Geometry


Date: 08/09/98 at 03:03:00
From: Simon Crosbie
Subject: Ellipse geometry

I wish to draw a line departing at a given angle from the long axis of 
an ellipse and bisecting the perimeter of the ellipse at right angles 
to the tangent at that point. How do I determine the distance along 
the long axis of an ellipse that this line must start from? 

I am a joiner who specializes in constructing special-shaped plantation 
shutters. A formula showing this relation would help me work out the 
precise timbers required to construct an ellipse. I have already worked 
out a formula which I use to draw an ellipse of a given width and 
height using two centers and a piece of string. The distance between 
the two centers is found with the expression:

   Center spacing = 2*Sqrt((Height/2)*(Height/2)*(Width/2))
   Length of string = Height + Center spacing

where Height is the length of the long axis of the ellipse and Width 
is the length of the short axis.


Date: 08/09/98 at 09:16:33
From: Doctor Jerry
Subject: Re: Ellipse geometry

Hi Simon,

Let's see if I understand your problem. You have an ellipse:

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

The numbers 2a and 2b are the lengths of the major and minor axes. The 
foci of the ellipse are at (-c,0) and (c,0), where c = sqrt(a^2-b^2). 
The length of the string is 2a.

I think 2a is what you have called width and 2b is what you have called 
height. I don't understand your formula for center spacing. I'd say 
that the center spacing (the distance between centers) is: 

   2c = 2*sqrt(a^2-b^2)

I'll use what I understand, as outlined in my first paragraph. You are 
seeking a point (X,0), where -a < X < a, on the major axis, from which 
a line at a specified angle d (I'll assume 0 < d < 90) will intercept 
the ellipse at right angles to the tangent.

There is a well-known formula for the equation of the tangent line to 
the ellipse at any point (x0,y0) of the ellipse. It is:

   a^2*y0*y + b^2*x*x0 = a^2*b^2

Assuming (x0,y0) is in the first quadrant, the slope of this line is
-b^2*x0/(a^2*y0). The slope of the line K perpendicular to this line is 
the negative reciprocal of this, namely, a^2*y0/(b^2*x0). This is the 
tangent of the angle d.

The equation of K is y-y0 = a^2*y0/(b^2*x0)(x-x0). Its x-intercept, 
which you want, is (set y=0):

   X = x0(1-b^2/a^2) = x0(a^2-b^2)/a^2 = x0*c^2/a^2

Since tan(d) = a^2*y0/(b^2*x0) and x0^2/a^2+y0^2/b^2 = 1, we can 
eliminate y0:

   1/x0^2 = (b^2/a^4)tan^2(d) + 1/a^2  

We now know x0 in terms of d. Put this into the equation for X and 
you're done.

Please check this out. Write back if I've made mistakes or been 
obscure.

- Doctor Jerry, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Conic Sections/Circles
High School Coordinate Plane Geometry
High School Geometry
High School Practical Geometry

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