Ellipse latus rectum

What the latus rectum is. Take the standard ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b>0$, foci at $(\pm c,0)$ where $c=\sqrt{a^2-b^2}$. Draw the vertical line through one focus — say $x=c$ — and mark where it cuts the ellipse. That chord, perpendicular to the major axis and passing through the focus, is the latus rectum. Because the ellipse is symmetric about the $x$-axis, this chord is automatically bisected by the axis, and because the ellipse is also symmetric about the $y$-axis, the chord through the other focus $(-c,0)$ has exactly the same length.

Why its length depends only on $a$ and $b$. Substituting $x=c$ into the ellipse equation isolates the $y$-values at that vertical slice:

where: $a$ is the semi-major axis, $b$ is the semi-minor axis, and $c=\sqrt{a^2-b^2}$ is the distance from the centre to a focus. Solving for $y$ and using $a^2-c^2=b^2$ gives $y=\pm\dfrac{b^2}{a}$ (worked in full below), so the two endpoints straddle the axis symmetrically and the chord length is twice that value.

The result, and how to read it. The length of the latus rectum is

where: $L$ is the latus rectum length, $b^2$ the square of the semi-minor axis, $a$ the semi-major axis. Since $b<a$ for any non-circular ellipse, $L=\dfrac{2b^2}{a}<2b$ — the latus rectum is always shorter than the minor axis, and it always lies strictly between $0$ and $2a$. A more flattened ellipse (larger eccentricity, smaller $b$ relative to $a$) has a shorter latus rectum, matching the intuition that an elongated ellipse pinches in sharply near its foci.

In terms of eccentricity. Since $b^2=a^2(1-e^2)$, the same length can be written as

where: $e$ is the eccentricity of the ellipse. This form is handy when a problem states $e$ directly, letting you skip finding $b^2$ first. Because the two latus recta (through each focus) are mirror images across the centre, both share this one value of $L$.

How $L$ tracks the ellipse's shape. Think of $e$ sweeping from $0$ (a circle) toward $1$ (an increasingly stretched ellipse) while $a$ stays fixed. At $e=0$, $L=2a(1-0)=2a$, matching the fact that a circle's "latus rectum" through its centre would simply be a diameter. As $e$ grows, the factor $(1-e^2)$ shrinks, so $L$ steadily falls — the chord through the focus gets pinched narrower the more elongated the ellipse becomes. This single formula therefore packages two separate facts you already know (the axis lengths and the eccentricity) into one number that tells you, at a glance, how "fat" or "thin" the ellipse is near its foci, which is why textbooks group it with eccentricity as one of the ellipse's basic descriptive properties rather than treating it as an isolated curiosity.

Problem. Find the length of the latus rectum of the ellipse $\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$, and verify the answer using eccentricity.

Step 1. Identify $a^2$ and $b^2$. The larger denominator sits under $x^2$, so $a^2=36 \Rightarrow a=6$, and $b^2=16$.

Step 2. Apply the formula. $L=\dfrac{2b^2}{a}=\dfrac{2\times16}{6}=\dfrac{32}{6}=\dfrac{16}{3}$.

Step 3. Cross-check via eccentricity. $c^2=a^2-b^2=36-16=20$, so $e^2=\dfrac{c^2}{a^2}=\dfrac{20}{36}=\dfrac{5}{9}$. Then $2a(1-e^2)=2(6)\left(1-\dfrac{5}{9}\right)=12\times\dfrac{4}{9}=\dfrac{48}{9}=\dfrac{16}{3}$ — matching Step 2 exactly.

Answer: the latus rectum has length $\boxed{\dfrac{16}{3}}$.

Competitive stretch

An ellipse has $a=10$ and latus rectum $12.8$. From $\dfrac{2b^2}{10}=12.8$, $b^2=64 \Rightarrow b=8$; then $c^2=100-64=36 \Rightarrow c=6$, giving $e=\dfrac{c}{a}=0.6$. Checking with $2a(1-e^2)=20(1-0.36)=20\times0.64=12.8$ confirms the values are consistent — a useful two-way habit for exams. Notice the pattern: whenever a problem gives you the latus rectum together with one of $a$, $b$, $c$ or $e$, it has effectively handed you the whole ellipse, since any two of the five quantities $a$, $b$, $c$, $e$, $L$ pin down the rest through the axis-focus relation $c^2=a^2-b^2$ and the two latus-rectum formulas above.

• Substituting $x=a$ (the vertex) instead of $x=c$ (the focus) — the latus rectum must pass through a focus, not a vertex.
• Confusing the ellipse formula $\dfrac{2b^2}{a}$ with the visually identical hyperbola formula — same expression, but check which conic's $a$ and $b$ you are using.
• Sign or placement errors in the endpoints — they are $\left(c,\pm\dfrac{b^2}{a}\right)$, not $\pm\dfrac{a^2}{b}$ or $\pm\dfrac{b}{a}$.
• Re-deriving $b^2$ from scratch when eccentricity is given directly, instead of using the shortcut $L=2a(1-e^2)$.
• Misreading which denominator is $a^2$ and which is $b^2$ when the equation is not already in the form $a>b$ — always compare the two denominators first; the larger one is $a^2$.

Ask the tutor: if you want to see step by step why substituting $x=c$ into the ellipse equation lands exactly on $y=\pm b^2/a$, or how to switch quickly between the $b$-based and eccentricity-based forms, ask the AI tutor to walk through it with you.