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]