﻿ Polar Coordinates

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.
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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

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 at the point (2c,0) (i.e., 2c is the distance from F1 to F2). Then r is the polar vector to the point P, and r-2ci  is the vector from F2 to P.
The ellipse definition implies that
 ||r||  +  ||r-2ci||   =   2a
Thus, r = ||r|| implies that r-2a = ||r-2ci||, so that
 ( r-2a) 2  =  ||r-2ci||2
However, r = á rcos( q) ,rsin(q) ñ implies that
 ( r - 2a)2
 =
 ( rcosq-2c)2  + ( r sinq)2
 ( r - 2a)2
 =
 r2cos2( q) - 2rccos( q) +  4c2 + r2sin2( q)
 r2 - 4ar + 4a2
 =
 r2 - 4rccos( q) + 4c2
Simplifying and solving for r yields
 -4ar + 4a2
 =
 -4rc cos( q) +4c2
 -ar + rccos( q)
 =
 -a2 + c2
 r( -a + ccos( q) )
 =
 -a2 + c2
 r
 =
-a2 + c2 -a + ccos( q)
If we let b2 = a2-c2 denote the square of the semi-minor axis, then
r =  b2 a - c cos( q)

Finally, let us divide by a to obtain
r =  b2/a 1-c/a cos( 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 7    Find the center, semi-major axis, semi-minor axis and foci of the ellipse
r =  3 1-0.5cos( q)
What is the Cartesian equation of the ellipse?

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.
To convert to x and y, we first multiply to get
 5r  -  4r cos(q) = 9
Thus, 5r = 4rcos( q) +9, so that
 25r2
 =
 ( 4rcos(q)  + 9)2
 25( x2+y2)
 =
 ( 4x + 9)2
Since ( 4x+9) 2 = 16x2+72x+81, we have
 25x2+25y2 = 16x2+72x+81
which yields 9x2-72x+25y2 = 81.

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.

Polar ellipses with F2 on the given axis and  F1 = (0,0)

positive x-axis:
r p 1-ecos( q)
positive y-axis:
r p 1-esin( q)
negative x-axis:
r p 1+ecos( q)
negative y-axis:
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.

EXAMPLE 8    Find the center, semi-major axis, semi-minor axis and foci of the ellipse
r = 16 5 + 3cos( q)

Solution: Let's divide the numerator and denominator by 5 to obtain
r = 16/5 1 + 3/5 cos( q)

Since p = 16/5 and e = 3/5, the formula p = a( 1-e2 ) implies that
a =  16/5 1-( 3/5) 2
=  5
As a result, b2 = ap implies that
b2 = 5 ·   16 5
= 16,        b = 4
Finally, the focus F1 is at the origin and the table above implies that F2 is on the negative y-axis. Since
 2ea = 2· (3/5) · 5 = 6
implies that F2 is at the point ( 0, -6).