<s>[INST] Finding the equation of an ellipse using eccentricity and directrix with focus at (0,0) The ellipse $\varepsilon$ has eccentricty $\frac{1}{2}$, focus $(0,0)$ and the line $x=-1$ as the corresponding directrix. Find the equation of $\varepsilon$. Find the other focus and directrix of $\varepsilon$.
I'm confused by this due to fact the focus is at $(0,0)$. As far as I was aware the center of the ellipse should be at $(0,0)$ so that the foci are at $(c,0)$ and $(-c,0)$, where $c^2=a^2-b^2$. The directrix corresponding to the focus will then be given by the equation $x=\frac{a^2}{c}$, and the eccentricity of the ellipse is $\frac{c}{a}$. Is there another set of equations I can use to determine the equation of the ellipse?
 [/INST] From your information:
\begin{align*}
  \frac{\sqrt{x^2+y^2}}{x+1} &= \varepsilon \\
  x^2+y^2 &= \varepsilon^2(x+1)^2 \\
  (1-\varepsilon^2)x^2-2\varepsilon^2 x+y^2 &= \varepsilon^2 \\
  \left[
    (1-\varepsilon^2)x^2-2\varepsilon^2 x+
    \frac{\varepsilon^4}{1-\varepsilon^2}
  \right]+
  y^2 &= \varepsilon^2+\frac{\varepsilon^4}{1-\varepsilon^2} \\
  \frac{(1-\varepsilon^2)^2}{\varepsilon^2}
  \left(
    x-\frac{\varepsilon^2}{1-\varepsilon^2}
  \right)^2+\frac{1-\varepsilon^2}{\varepsilon^2} y^2 &= 1 \\
  a &= \frac{\varepsilon}{1-\varepsilon^2} \\
  b &= \frac{\varepsilon}{\sqrt{1-\varepsilon^2}} \\
  c &= \frac{\varepsilon^2}{1-\varepsilon^2} 
\end{align*}
The other focus is
$$(2c,0)=
\left( \frac{2\varepsilon^2}{1-\varepsilon^2},0 \right)$$
The other directrix is
$$x=2c+1$$
$$x=\frac{1+\varepsilon^2}{1-\varepsilon^2}$$

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