Polar Equation of an Ellipse

Date: 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