First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is [latex]\frac{1}{3}[/latex].

[latex]\begin{align} r&=\frac{5}{3+3 \cos \theta } \\ &=\frac{5\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+3\left(\frac{1}{3}\right)\cos \theta } \\ r&=\frac{\frac{5}{3}}{1+\cos \theta } \end{align}[/latex]

Because [latex]e=1[/latex], we will graph a parabola with a focus at the origin. The function has a [latex]\cos \text{ }\theta[/latex], and there is an addition sign in the denominator, so the directrix is [latex]x=p[/latex].

[latex]\begin{align}\frac{5}{3}&=ep\\ \frac{5}{3}&=\left(1\right)p\\ \frac{5}{3}&=p\end{align}[/latex]

The directrix is [latex]x=\frac{5}{3}[/latex].

Plotting a few key points as in the table below will enable us to see the vertices.

	A	B	C	D
[latex]\theta[/latex]	[latex]0[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\pi[/latex]	[latex]\frac{3\pi }{2}[/latex]
[latex]r=\frac{5}{3+3\text{ }\cos \text{ }\theta }[/latex]	[latex]\frac{5}{6}\approx 0.83[/latex]	[latex]\frac{5}{3}\approx 1.67[/latex]	undefined	[latex]\frac{5}{3}\approx 1.67[/latex]

Analysis of the Solution

We can check our result with a graphing utility.

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is [latex]\frac{1}{2}[/latex].

[latex]\begin{align} r&=\frac{8}{2 - 3\sin \theta } \\ &=\frac{8\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-3\left(\frac{1}{2}\right)\sin \theta } \\ r&=\frac{4}{1-\frac{3}{2} \sin \theta } \end{align}[/latex]

Because [latex]e=\frac{3}{2},e>1[/latex], so we will graph a hyperbola with a focus at the origin. The function has a [latex]\sin \text{ }\theta[/latex] term and there is a subtraction sign in the denominator, so the directrix is [latex]y=-p[/latex].

[latex]\begin{align}4&=ep \\ 4&=\left(\frac{3}{2}\right)p \\ 4\left(\frac{2}{3}\right)&=p \\ \frac{8}{3}&=p \end{align}[/latex]

The directrix is [latex]y=-\frac{8}{3}[/latex].

Plotting a few key points as in the table below will enable us to see the vertices.

	A	B	C	D
[latex]\theta[/latex]	[latex]0[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\pi[/latex]	[latex]\frac{3\pi }{2}[/latex]
[latex]r=\frac{8}{2 - 3\sin \theta }[/latex]	[latex]4[/latex]	[latex]-8[/latex]	[latex]4[/latex]	[latex]\frac{8}{5}=1.6[/latex]

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is [latex]\frac{1}{5}[/latex].

[latex]\begin{align}r&=\frac{10}{5 - 4\cos \theta } \\ &=\frac{10\left(\frac{1}{5}\right)}{5\left(\frac{1}{5}\right)-4\left(\frac{1}{5}\right)\cos \theta } \\ r&=\frac{2}{1-\frac{4}{5}\cos \theta } \end{align}[/latex]

Because [latex]e=\frac{4}{5},e<1[/latex], so we will graph an ellipse with a focus at the origin. The function has a [latex]\text{cos}\theta[/latex], and there is a subtraction sign in the denominator, so the directrix is [latex]x=-p[/latex].

[latex]\begin{align}2&=ep \\ 2&=\left(\frac{4}{5}\right)p \\ 2\left(\frac{5}{4}\right)&=p \\ \frac{5}{2}&=p \end{align}[/latex]

The directrix is [latex]x=-\frac{5}{2}[/latex].

Plotting a few key points as in the table below will enable us to see the vertices.

	A	B	C	D
[latex]\theta[/latex]	[latex]0[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\pi[/latex]	[latex]\frac{3\pi }{2}[/latex]
[latex]r=\frac{10}{5 - 4\text{ }\cos \text{ }\theta }[/latex]	[latex]10[/latex]	[latex]2[/latex]	[latex]\frac{10}{9}\approx 1.1[/latex]	[latex]2[/latex]

Analysis of the Solution

We can check our result using a graphing utility.