Ellipses in Polar Coordinates

Let's suppose that 2 ''nails'' are driven into a board at points F1 and F2, and suppose that the ends of a string of length 2a is attached to the board at points F1 and F2. If the string is pulled tight around a pencil's tip, then the points P traced by the pencil as it moves within the string form an ellipse.

Specifically, an ellipse is the locus of all points P such that
| PF1| +| PF2| = 2a
where | PF1| and | PF2| denote distances from P to F1 and F2, respectively. The points F1 and F2 are called the foci of the ellipse, and the distance a is called the semi-major axis.

Let's use this definition of an ellipse to derive its representation in polar coordinates. To begin with, let's assume that F1 is at the origin and that F2 is on the positive real axis. Then the distance |PF1| is the same as the r in polar coordinates

If we let d = | PF2| , then r+d = 2a, which implies that d = 2a-r.

Now let us let 2c denote the distance from F1 to F2, and let ( x,y) be the coordinates of P

Then the Pythagorean theorem implies that

( 2c-x) 2+y2  =  d2
4c2-4cx+x2+y2  =  d2
Thus, d = 2a-r and x = rcos( q) imply that
4c2-4crcos( q) +r2
=
( 2a-r) 2
4c2-4crcos( q) +r2
=
4a2-4ar+r2
4ar-4crcos( q)
=
4a2-4c2
r( a-ccos( q) )
=
a2-c2
r
=
 b2
a-ccos( q)
where b2 = a2-c2 is the square of the semi-minor axis.

Finally, let us divide by a to obtain
r =  b2/a
1-c/acos( q)
Usually, we let e = c/a and let p = b2/a, where e is called the eccentricity of the ellipse and p is called the parameter. It follows that 0 £ e < 1 and p > 0, so that an ellipse in polar coordinates with one focus at the origin and the other on the positive x-axis is given by
r =  p
1-ecos( q)
Moreover, b2 = a2-c2 implies that
p = a æ
è 		 b2
a2
 ö
ø 		= a æ
è 		 a2-c2
a2
 ö
ø 		= a æ
è 		1-  c2
a2
 ö
ø 		,
which in turn implies that p = a( 1-e2) .

       

EXAMPLE 8    Find the center, semi-major axis, semi-minor axis and foci of the ellipse
r =  3
1-0.5cos( q)

Solution: Since p = 3 and e = 0.5, the formula p = a( 1-e2 ) implies that
a =  3
1-( 0.5) 2
 = 4
As a result, b2 = ap implies that
b2 = 4·3 = 12,        b = Ö12 = 2Ö3
Finally, the focus F1 is at the origin and the calculation
2ea = 2·0.5·4 = 4
implies that F2 is at the point ( 4,0) on the x-axis.

       

Finally, had F2 been placed on the positive y-axis, the negative x-axis, or the negative y-axis, the polar equation of the ellipse would be modified by replacing cos( q) with sin( q) , -cos( q) , or -sin( q) , respectively.
r =  p
1-ecos( q)
r =  p
1-esin( q)
r =  p
1+ecos( q)
r =  p
1+esin( q)

Indeed, the most general form of an conic with parameter p and eccentricity e > 0 is
r =  p
1-ecos( q-q0)
(3)
where it can be shown that a = ecos( q0) and b = esin( q0) . It follows that the conic is symmetric about the line at angle q0 to the x-axis.