# Finding the second focus of an ellipse and its directrix

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?

• math.stackexchange.com/questions/49517/… Commented Nov 8, 2016 at 12:42
• Thank you, this is what I have mentioned in part B) but this does not help for a focus of (0,0) as it seems to imply the other focus is (0,0) with a directrix of x=(a^2)/0 which is wrong. Thanks anyway! Commented Nov 8, 2016 at 12:59
• Your equation for the ellipse is clearly incorrect: both squared terms have the same coefficient, so the eccentricity of the conic described by this equation can’t be $\frac12$. Moreover, the center that one computes from this equation lies between the given focus and directrix, so the equation can’t be correct. What you’ve found is the equation of a circle that’s tangent to the directrix with center on the $x$-axis.
– amd
Commented Jul 18, 2017 at 20:15

## 3 Answers

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 $x=2c+1$.

In the different formulas you have different coordinate systems. The coordinate system used by your problem statement and presumably the equation you found has one focus in the origin. The symmetric focus coordinates you quote assume that the origin is in the center of symmetry, and furthermore that the first coordinate axis agrees with the major axis of the ellipse.

There are different ways to go about this. If you have computed $c$, you can just use the quoted formula to conclude that the distance between the center and the focus is $c$, so the distance between foci is $2c$, so if you know the direction where the other focus has to lie, you can easily compute its position.

If you don't have that $c$ yet, you could also start by finding the center of the ellipse. As your ellipse is axis-aligned, this is easy to do by completing the square. Once you have the center, then the second focus and directrix are what you get by reflecting the first focus and directrix in that center.

Or you could take the equational approach. You know that things have to be symmetric, so as the first directrix is one unit left of the first focus, the second directrix is one unit right of the second focus. Assume the second focus at $(f,0)$ and the second directrix at $x=f+1$. Find a second formula of the ellipse, depending on $f$, and then make the choose $f$ in such a way that the two formulas agree.

• Thanks, I have tried to complete the square to find the center of the ellipse but have ended up with (x+1/3)^2 + y^2=4/9 which is just a circle center (-1/3,0), radius 2/3 which I know is wrong. Can you see where I've gone wrong? Commented Nov 9, 2016 at 14:20
• @harry55: I take it that my answer to your separate question on this should resolve this issue here.
– MvG
Commented Nov 9, 2016 at 15:25

In order to find both x- coordinates, set $y=0$ in the equation you got and solve for two $x$ values.

$$3x^2 + 2x -1 = (3 x -1) (x+1) = 0$$

$$(x= \frac13, \, x=-1)$$

Other focus is found.

Now the foci and directrix centre points have same midpoint. Find other directrix.