Sketch the graph of each conic section. State the foci, vertices, and eccentricity. 3/(2-cos(theta))?

Answer 1

r = 3/(2 - cosθ)

r(2 - cosθ) = 3

2r - rcosθ = 3

2r - x = 3

r = (x + 3) / 2

r² = (x + 3)² / 4

x² + y² = (x + 3)² / 4

4x² + 4y² = (x + 3)² = x² + 6x + 9

3x² - 6x - 9 + 4y² = 0

3(x² - 2x - 3) + 4y² = 0

3((x - 1)² - 4) + 4y² = 0

(x - 1)² - 4 + 4y²/3 = 0

(x - 1)² + 4y²/3 = 4

(x - 1)²/4 + y²/3 = 1

which is an ellipse and you should be able to finish it by
comparing it with the standard ellipse

Answer 2

The equation of the conic section is

r = 3/(2 - cosθ)
_________________

This is the equation of an ellipse.

Let's put this in a more helpful form. Divide numerator and denominator by 2.

r = (3/2) / (1 - ½cosθ)

This is the standard polar form of the equation of a conic section with one focus at the pole (origin). We wanted the first term in the denominator to be a one.

We can find the eccentricity ε in the denominator.

ε = 1/2

Since the eccentricity is less than one, the conic section is an ellipse.

Cosine tells us that the major axis is horizontal. (Sine would mean the major axis was vertical.) The minus sign tells us the second focus is on the positive axis. (A plus sign would mean the second focus was on the negative axis.)

The numerator is the semi-latus rectum l.

l = a(1 - ε²)
a = l/(1 - ε²) = (3/2) / (1 - ½²) = (3/2) / (3/4) = 2
a² = 4

l = b²/a
b² = al = 2*(3/2) = 3

c² = a² - b² = 4 - 3 = 1
c = 1

One focus is at the origin (0,0). The other is at
(0+2c,0) = (2,0).

The foci are
(0,0) and (2,0)

The center of the ellipse is the midpoint of the foci (1, 0).
(h,k) = (1, 0)

The vertices are (h-a,k) and (h+a,k) or (1-2, 0) and (1+2, 0).

The vertices are
(-1, 0) and (3, 0)