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?
Best Answer
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 $$x=2c+1$$ $$x=\frac{1+\varepsilon^2}{1-\varepsilon^2}$$