For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, [latex]\frac{1}{c}[/latex], where [latex]c[/latex] is that constant.

1. Multiply the numerator and denominator by [latex]\frac{1}{3}[/latex].
   [latex]\begin{align}r&=\frac{6}{3+2\sin \theta }\cdot \frac{\left(\frac{1}{3}\right)}{\left(\frac{1}{3}\right)} \\ &=\frac{6\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right)\sin \theta } \\ r&=\frac{2}{1+\frac{2}{3}\sin \theta} \end{align}[/latex]

   Because [latex]\sin \text{ }\theta [/latex] is in the denominator, the directrix is [latex]y=p[/latex]. Comparing to standard form, note that [latex]e=\frac{2}{3}[/latex]. Therefore, from the numerator,

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

   Since [latex]e<1[/latex], the conic is an ellipse. The eccentricity is [latex]e=\frac{2}{3}[/latex] and the directrix is [latex]y=3[/latex].

2. Multiply the numerator and denominator by [latex]\frac{1}{4}[/latex].
   [latex]\begin{align} r&=\frac{12}{4+5 \cos \theta }\cdot \frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{4}\right)} \\ &=\frac{12\left(\frac{1}{4}\right)}{4\left(\frac{1}{4}\right)+5\left(\frac{1}{4}\right)\cos \theta } \\ r&=\frac{3}{1+\frac{5}{4}\cos \theta } \end{align}[/latex]

   Because [latex]\text{ cos}\theta [/latex] is in the denominator, the directrix is [latex]x=p[/latex]. Comparing to standard form, [latex]e=\frac{5}{4}[/latex]. Therefore, from the numerator,

   [latex]\begin{align}3&=ep \\ 3&=\frac{5}{4}p \\ \left(\frac{4}{5}\right)3&=\left(\frac{4}{5}\right)\frac{5}{4}p \\ \frac{12}{5}&=p \end{align}[/latex]

   Since [latex]e>1[/latex], the conic is a hyperbola. The eccentricity is [latex]e=\frac{5}{4}[/latex] and the directrix is [latex]x=\frac{12}{5}=2.4[/latex].

3. Multiply the numerator and denominator by [latex]\frac{1}{2}[/latex].
   [latex]\begin{align} r&=\frac{7}{2 - 2 \sin \theta }\cdot \frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)} \\ &=\frac{7\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-2\left(\frac{1}{2}\right) \sin \theta } \\ r&=\frac{\frac{7}{2}}{1-\sin \theta } \end{align}[/latex]

   Because sine is in the denominator, the directrix is [latex]y=-p[/latex]. Comparing to standard form, [latex]e=1[/latex]. Therefore, from the numerator,

   [latex]\begin{align}\frac{7}{2}&=ep\\ \frac{7}{2}&=\left(1\right)p\\ \frac{7}{2}&=p\end{align}[/latex]

   Because [latex]e=1[/latex], the conic is a parabola. The eccentricity is [latex]e=1[/latex] and the directrix is [latex]y=-\frac{7}{2}=-3.5[/latex].

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.

Figure 3

Analysis of the Solution

We can check our result with a graphing utility.

Figure 4

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.

Figure 5

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.

Figure 6

Analysis of the Solution

We can check our result using a graphing utility.

Figure 7. [latex]r=\frac{10}{5 – 4 \cos \theta }[/latex] graphed on a viewing window of [latex]\left[-3,12,1\right][/latex] by [latex]\left[-4,4,1\right],\theta \text{min =}0[/latex] and [latex]\theta \text{max =}2\pi [/latex].

The directrix is [latex]y=-p[/latex], so we know the trigonometric function in the denominator is sine.

Because [latex]y=-2,-2<0[/latex], so we know there is a subtraction sign in the denominator. We use the standard form of

[latex]r=\frac{ep}{1-e \sin \theta }[/latex]

and [latex]e=3[/latex] and [latex]|-2|=2=p[/latex].

Therefore,

[latex]\begin{align}r&=\frac{\left(3\right)\left(2\right)}{1 - 3 \sin \theta } \\ r&=\frac{6}{1 - 3 \sin \theta } \end{align}[/latex]

Because the directrix is [latex]x=p[/latex], we know the function in the denominator is cosine. Because [latex]x=4,4>0[/latex], so we know there is an addition sign in the denominator. We use the standard form of

[latex]r=\frac{ep}{1+e \cos \theta }[/latex]

and [latex]e=\frac{3}{5}[/latex] and [latex]|4|=4=p[/latex].

Therefore,

[latex]\begin{align} r&=\frac{\left(\frac{3}{5}\right)\left(4\right)}{1+\frac{3}{5}\cos \theta } \\ &=\frac{\frac{12}{5}}{1+\frac{3}{5}\cos \theta } \\ &=\frac{\frac{12}{5}}{1\left(\frac{5}{5}\right)+\frac{3}{5}\cos \theta } \\ &=\frac{\frac{12}{5}}{\frac{5}{5}+\frac{3}{5}\cos \theta } \\ &=\frac{12}{5}\cdot \frac{5}{5+3\cos \theta } \\ r&=\frac{12}{5+3\cos \theta } \end{align}[/latex]

We will rearrange the formula to use the identities [latex] r=\sqrt{{x}^{2}+{y}^{2}},x=r\cos \theta ,\text{and }y=r\sin \theta [/latex].

[latex]\begin{align}&r=\frac{1}{5 - 5\sin \theta } \\ &r\cdot \left(5 - 5\sin \theta \right)=\frac{1}{5 - 5\sin \theta }\cdot \left(5 - 5\sin \theta \right)&& \text{Eliminate the fraction}. \\ &5r - 5r\sin \theta =1&& \text{Distribute}. \\ &5r=1+5r\sin \theta && \text{Isolate }5r. \\ &25{r}^{2}={\left(1+5r\sin \theta \right)}^{2}&& \text{Square both sides}. \\ &25\left({x}^{2}+{y}^{2}\right)={\left(1+5y\right)}^{2}&& \text{Substitute }r=\sqrt{{x}^{2}+{y}^{2}}\text{ and }y=r\sin \theta . \\ &25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2}&& \text{Distribute and use FOIL}. \\ &25{x}^{2}-10y=1&& \text{Rearrange terms and set equal to 1}. \end{align}[/latex]