Eccentricity, Foci, Directrices and the Latus Rectum

Locating the foci and directrices from a and b

Once an ellipse is written as $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b>0$, every other feature of the curve — the two foci, the two directrices, and the eccentricity itself — can be recovered directly from $a$ and $b$, without going back to the defining focus-directrix picture each time. From $b^2 = a^2(1-e^2)$ we can solve for the eccentricity:

Because the ellipse is symmetric about the y-axis as well as the x-axis, there isn't just one focus-directrix pair — there are two, mirror images of each other in the centre:

If instead the y-denominator is the larger one ($0<a<b$, major axis vertical), the same logic gives foci $(0,\pm be)$, directrices $y=\pm b/e$, and $e=\sqrt{(b^2-a^2)/b^2}$ — just swap the roles of the axis carrying the bigger denominator.

The latus rectum

A chord through a focus, perpendicular to the major axis, is called a latus rectum (there are two, one through each focus). To find its length, substitute $x=ae$ into the ellipse equation: $\dfrac{(ae)^2}{a^2} + \dfrac{y^2}{b^2}=1 \Rightarrow y^2 = b^2(1-e^2) = b^4/a^2$ (using $b^2=a^2(1-e^2)$), so $y = \pm b^2/a$. The two endpoints are $\left(ae, \dfrac{b^2}{a}\right)$ and $\left(ae, -\dfrac{b^2}{a}\right)$, giving

This single number is a compact way to describe how "wide" the ellipse is at the focus — a useful check figure alongside $a$, $b$, and $e$.

Worked Example

Problem. For the ellipse $\dfrac{x^2}{169} + \dfrac{y^2}{144} = 1$, find the eccentricity, the coordinates of the foci, the equations of the directrices, and the length of the latus rectum.

Solution. Here $a^2 = 169 \Rightarrow a=13$ and $b^2=144 \Rightarrow b=12$; since $a>b$ the major axis is along the x-axis.

$e = \sqrt{\frac{a^2-b^2}{a^2}} = \sqrt{\frac{169-144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}.$ …