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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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).