Ellipse GeometryDate: 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/ |
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