My principle question is how do I find the other focus of the ellipse if one is (0,0)??
For context I have done the following:
A) The ellipse $E$ has eccentricity $\frac12$, focus $(0,0)$ with the line $x=-1$ as the corresponding directrix. Find an equation for $E$.
I have attempted this using $\frac{\lvert PF\rvert}{\lvert Pl\rvert}=\frac12$.
Where $P$ is an arbitrary point, $F$ is the focus and $l$ is the directrix and have ended up with the equation $$\tfrac34x^2 + \tfrac34y^2 + \tfrac12x -\tfrac14=0$$ It is the next part I am having difficulty with.
B) Find the other focus and directrix of E.
I know that in the general equation of an ellipse, the foci are given by $(c,0)$ and $(-c,0)$ with $c^2=a^2-b^2$ but I am not sure how to use this information when my equation for $E$ is not in the general form and the focus is $(0,0)$. How can it be $(0,0)$ if they're symmetric about the origin??
I tried to put my equation for $E$ in the general form but got stuck so perhaps my attempt is wrong?
Best Answer
The foci and directrices are symmetic with respect to the minor axis, so once you have the center, you can find the other focus and directrix by reflection in the line through this point parallel to the known directrix. The following solution doesn’t require first finding the equation of the ellipse.
Let $D$ be the intersection of the directrix and its perpendicular through the focus $F$ (i.e., with the major axis of the ellipse). One vertex $A$ of the ellipse is between these two points. For this vertex, $|A-F|=e|D-A|$, so $A={1\over1+e}F+{e\over1+e}D$. For the other vertex $A'$, we have $|A'-F|=e|A'-D|$, from which $A'={1\over1-e}F-{e\over1-e}D$. The center of the ellipse is their midpoint $$C=\frac12(A+A')={1\over1-e^2}F-{e^2\over1-e^2}D.$$ In this problem, $F=(0,0)$ and $D=(-1,0)$, so the center of the ellipse is at $${1\over1-\left(\frac12\right)^2}(0,0)-{\left(\frac12\right)^2\over1-\left(\frac12\right)^2}(-1,0)=-\frac14\cdot\frac43\cdot(-1,0)=\left(\frac13,0\right).$$ From this, we easily find that the other focus is $\left(\frac23,0\right)$ with corresponding directrix $x=\frac53$.
If you have the equation of the conic in hand, you can find its center by differentiating and setting the partial derivatives to zero, i.e., $$\frac32x+\frac12=0\\ \frac32y=0$$ which has solution $x=-\frac13$, $y=0$. This point is between the given focus and directrix, so as I mention in my comment to your question, this means that the equation that you’ve come up with is incorrect. That aside, once you’ve found the correct center point $(c,0)$, the other focus is the reflection in this point, namely, $(2c,0)$, with corresponding directrix $x=2c+1$.