Answer
See image:
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Work Step by Step
The polar equation for a conic with eccentricity $e$ is $\displaystyle \quad r=\frac{ke}{1+e\cos\theta},$
where
$x=k \gt 0$ is the vertical directrix, and $\left\{\begin{array}{ll}
e=1 & \Rightarrow\text{parabola}\\
e \lt 1 & \Rightarrow\text{ellipse}\\
e \gt 1 & \Rightarrow\text{hyperbola}
\end{array}\right.$
$\displaystyle \frac{25\div 10}{(10-5\cos\theta)\div 10}=\frac{\frac{5}{2}}{1-(\frac{1}{2})\cos\theta}=\frac{-5(-\frac{1}{2})}{1+(-\frac{1}{2})\cos\theta}$
So, $e=\displaystyle \frac{1}{2},\qquad \Rightarrow\text{ellipse}$
and $\quad k=5 \Rightarrow x=-5$ is a directrix
The origin is the left focus.
$ r=\displaystyle \frac{a(1-e^{2})}{1+e\cos\theta}\quad$ is the polar equation of an ellipse.
$a(1-e^{2})=\displaystyle \frac{5}{2}$
$a(1-\displaystyle \frac{1}{4})=\frac{5}{2}$
$a\displaystyle \cdot\frac{3}{4}=\frac{5}{2}$
$a=\displaystyle \frac{10}{3}$
$c=ea=\displaystyle \frac{1}{2}\cdot\frac{10}{3}=\frac{5}{3}$
The center of the ellipse is $\displaystyle \frac{5}{3}$ units right of the focus at the origin.
In polar coordinates, at $(\displaystyle \frac{5}{3},0).$
The vertices are at $\displaystyle \frac{10}{3}$ units left/right of the center; in polar coordinates,
$(\displaystyle \frac{5}{3}+\frac{10}{3},0)=(5,0)$ and
$(\displaystyle \frac{5}{3}-\frac{10}{3},0)=(-\frac{5}{3},0)$ or $(\displaystyle \frac{5}{3},\pi)$
