In order to find the polar equation of the ellipse, first let's find the rectangular equation.
Since the directrix is a vertical line, the ellipse is horizontal, and the model equation is:
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]Where the center is located at (h, k), the directrix is x = -a/e and the eccentricity is e = c/a.
So, if the eccentricity is e = 1/5 and the directrix is x = -7, we have:
[tex]\begin{gathered} \frac{c}{a}=\frac{1}{5}\rightarrow a=5c\\ \\ -\frac{a}{e}=-7\\ \\ \frac{a}{\frac{c}{a}}=7\\ \\ \frac{a^2}{c}=7\\ \\ \frac{25c^2}{c}=7\\ \\ 25c=7\\ \\ c=\frac{7}{25}\\ \\ a=5\cdot\frac{7}{25}=\frac{7}{5} \end{gathered}[/tex]Now, let's calculate the value of b with the formula below:
[tex]\begin{gathered} c^2=a^2-b^2\\ \\ \frac{49}{625}=\frac{49}{25}-b^2\\ \\ b^2=\frac{25\cdot49}{625}-\frac{49}{625}\\ \\ b^2=\frac{24\cdot49}{625}\\ \\ b^2=\frac{1176}{625} \end{gathered}[/tex]Assuming h = 0 and k = 0, the rectangular equation is:
[tex]\frac{x^2}{\frac{49}{25}}+\frac{y^2}{\frac{1176}{625}}=1[/tex]Now, to convert to polar form, we can do the following steps:
[tex]\begin{gathered} \frac{25x^2}{49}+\frac{625y^2}{1176}=1\\ \\ 600x^2+625y^2=1176\\ \\ 600(r\cos\theta)^2+625(r\sin\theta)^2=1176\\ \\ 600r^2\cos^2\theta+625r^2\sin^2\theta=1176\\ \\ r^2(600\cos^2\theta+625\sin^2\theta)=1176\\ \\ r^2=\frac{1176}{600\cos^2\theta+625\sin^2\theta}\\ \\ r=\sqrt{\frac{1176}{600\cos^2\theta+625\sin^2\theta}}\\ \\ r=\sqrt{\frac{1176}{600+25\sin^2\theta}} \end{gathered}[/tex]Another way of writing this equation in polar form is:
[tex]r=\frac{ep}{1+\sin^2\theta}[/tex]Where p is the distance between the focus and the directrix.
Since the foci are located at (±c, 0) = (±7/25, 0) and the directrix is x = -7, the distance is:
[tex]p=7-\frac{7}{25}=\frac{175}{25}-\frac{7}{25}=\frac{168}{25}[/tex]So the equation is:
[tex]\begin{gathered} r=\frac{\frac{1}{5}\cdot\frac{168}{25}}{1+\sin^2\theta}\\ \\ r=\frac{\frac{168}{125}}{1+\sin^2\theta}\\ \\ r=\frac{1.344}{1+\sin^2\theta} \end{gathered}[/tex]