Steven Dutch, Professor Emeritus, Natural and Applied Sciences,University of Wisconsin - Green Bay
The equation of an ellipse as given in most references is:
x2/a2 + y2/b2 = 1 y
It would be useful to represent the equation in polar coordinates. To do this, we note that in polar coordinates, x = r cos A and y = r sin A, where A is the azimuth of point (x,y). We plug those formulas into the equation for an ellipse and get:
r2cos2A/a2 + r2sin2A/b2 = 1
We solve for r and get:
(1/a2) cos2A + (1/b2) sin2A = 1/r2
From elementary trigonometry, we recall the double-angle formulas:cos 2A = cos2A - sin2A = 1 - 2 sin A = 2 cos A - 1We can rearrange the terms to get:
We can substitute functions of 2A into the polar equation for an ellipse to get:
(1/a2)(cos 2A + 1)/2 + (1/b2)(1 - cos 2A)/2 = 1/r2
We can group terms to obtain:
[(1/b2) + (1/a2)]/2 + {[(1/b2) - (1/a2)]/2}cos 2A = 1/r2
This is very reminiscent of the development of the Mohr Circle for stress. We could construct a graph whose x-dimension represents the inverse of the radius squared. Next we draw a circle passing through 1/a^2 and 1/b^2 (note that if a is the major axis, 1/b^2 is larger). If we construct a radius of the circle that makes an angle 2A with the x-axis (measured from the 1/a^2 end of the diameter), then the x-distance of the
The question now is, what does the y-coordinate represent? The y-coordinate value is
{[(1/b2) - (1/a2)]/2}sin 2A or,
[[(1/b2) - (1/a2)]/2}sin A cos A
We recall that x = r cos A and y = r sin A, or x/r = cos A and y/r = sin A. We substitute these expressions for the trigonometric terms to get:
{[(1/b2) - (1/a2)]/2} xy/r
Now the equation for a circle with center (p,0) and radius r is:
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Created
13 May 1999, Last Update 13 May 1999