Polar Equation of an EllipseDate: 12/16/95 at 22:30:56 From: Anonymous Subject: Geometry Find the polar equation of the ellipse with eccentricity 3/4, one focus at the pole, and the corresponding directrix perpendicular to the polar axis, through the point with polar coordinates (8,pi). Date: 5/30/96 at 14:59:10 From: Doctor Charles Subject: Re: Geometry The distance of a point on the ellipse from the directrix is (x-8) and its distance from the origin is sqrt(x^2+y^2). So by the definition of the ellipse: (x-8)^2=e^2*(x^2+y^2) e is the eccentricity So in polar coordinates: (r cos(t) - 8)^2 = r^2 * e^2 r^2 (cos^2 (t) - e^2) - r * 16 cos(t) + 64 = 0 This quadratic can be solved for r in terms of t. (apply the formula) -Doctor Charles, The Math Forum Date: 5/30/96 at 15:0:12 From: Doctor Anthony Subject: Re: Geometry Let S be the focus of the conic and KL its directrix. SZ is the perpendicular from S on KL, and SZ is 180 degrees from the initial line, with S as the pole of coordinates. P=(r,theta) is any point on the conic and ST is the semi-latus rectum of length L. The eccentricity of the conic is e, and TR is the perpendicular from T to KL. PN is the ordinate of P, and PM is the perpendicular from P to KL. From the definition of a conic, ST = e.TR, that is, L = e.TR so TR = L/e From the diagram SZ = TR = L/e Now SN = r.cos(theta) and SP = r = e.PM Therefore PM = NZ = r/e Also SZ = NZ - NS Therefore L/e = r/e - r.cos(theta) So the polar equation is L/r = 1 - e.cos(theta) To find the polar equation of the directrix, using same diagram let X = (r,theta) be a point on the directrix KL SZ = SX.cos(pi-theta) = -SX.cos(theta) = -r.cos(theta) But SZ = L/e, therefore L/e = -r.cos(theta) is the equation of the directrix. We are given that e = 3/4 and that when theta=pi, r=8 So L/(3/4) = -8.cos(pi). From this L = (3/4)(8) = 6 The equation of our ellipse is: 6/r = 1 - (3/4).cos(theta) -Doctor Anthony, The Math Forum |
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