7-09 Polar Graphs of Conics

Identify Type of Conic

Identify the type of conic of $r = \frac{5}{2+4 \sin θ}$.

Solution

All the forms of a conic equation start with a 1 in the denominator. To get that, multiply the numerator and denominator by $\frac{1}{2}$.

$$r = \frac{5}{2+4 \sin θ}$$

$$r = \frac{\frac{5}{2}}{1 + 2 \sin θ}$$

The coefficient of the trig function is e. In this case, e = 2 > 1 so it is a hyperbola. The denominator has a $+ \sin θ$ which looks like $r = \frac{ep}{1+e \sin θ}$. So the conic is a hyperbola with horizontal directrix above the pole.

Identify Type of Conic

Identify the type of conic of $r = \frac{5}{3-\cos θ}$ and then graph it.

Solution

All the forms of a conic equation start with a 1 in the denominator. To get that, multiply the numerator and denominator by $\frac{1}{3}$.

$$r = \frac{5}{3-\cos θ}$$

$$r = \frac{\frac{5}{3}}{1 - \frac{1}{3} \cos θ}$$

The coefficient of the trig function is e. In this case, $e = \frac{1}{3} < 1$ so it is a ellipse. The denominator has a $- \cos θ$ which looks like $r = \frac{ep}{1-e \cos θ}$. So the conic is a ellipse with vertical directrix left of the pole.

Graph the ellipse by making a table of values.

r	2.5	2.46	2.34	2.18	2	1.82	1.67	1.53	1.43	1.35	1.29	1.26	1.25
θ	0	$\frac{π}{12}$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{5π}{12}$	$\frac{π}{2}$	$\frac{7π}{12}$	$\frac{2π}{3}$	$\frac{3π}{4}$	$\frac{5π}{6}$	$\frac{11π}{12}$	$π$

r	1.26	1.29	1.35	1.43	1.53	1.67	1.82	2	2.18	2.34	2.46	2.5
θ	$\frac{13π}{12}$	$\frac{7π}{6}$	$\frac{5π}{4}$	$\frac{4π}{3}$	$\frac{17π}{12}$	$\frac{3π}{2}$	$\frac{19π}{12}$	$\frac{5π}{3}$	$\frac{7π}{4}$	$\frac{11π}{6}$	$\frac{23π}{12}$	$2π$

Find the Polar Equation of a Conic

Find the polar equation of the parabola whose focus is the pole and directrix is the line $y = 3$.

Solution

Since the focus is the origin and the directrix is y = 3, the directrix is horizontal and above the pole. Thus, our equation is

$$r = \frac{ep}{1 + e \sin θ}$$

Since it is a parabola, e = 1. p is the distance from the focus (0, 0) to the directrix y = 3, so p = 3. Fill those in and the equation becomes

$$r = \frac{1(3)}{1 + 1 \sin θ}$$

$$r = \frac{3}{1 + \sin θ}$$

Find the Polar Equation of a Conic

Find the polar equation of the ellipse whose focus is the pole and has vertices (2, 0) and (8, π).

Solution

Draw a quick sketch of the points and draw the primary axis of the ellipse. Remember the points are in polar format.

The directrix will be to the right and the ellipse will bend away from it. So the directrix is vertical and to the right of the origin.

$$r = \frac{ep}{1 + e \cos θ}$$

The center is halfway between the two vertices. a is the distance from the center to vertex, and c is the distance from the center to the focus. Remember from before that $e = \frac{c}{a}$. In this case, a = 5 and c = 3, so $e = \frac{3}{5}$. Fill this in and the equation becomes

$$r = \frac{\frac{3}{5}p}{1 + \frac{3}{5} \cos θ}$$

Multiply numerator and denominator by 5 to remove fractions.

$$r = \frac{3p}{5 + 3 \cos θ}$$

Now plug in a vertex point such as (2, 0) and solve for p.

$$2 = \frac{3p}{5 + 3 \cos 0}$$

$$2 = \frac{3p}{8}$$

$$16 = 3p$$

$$\frac{16}{3} = p$$

Plug this into $r = \frac{3p}{5 + 3 \cos θ}$ and simplify to get the final equation.

$$r = \frac{3\left(\frac{16}{3}\right)}{5 + 3 \cos θ}$$

$$r = \frac{16}{5 + 3 \cos θ}$$