Hyperbola latus rectum

Definition. The latus rectum is a chord through a focus, drawn perpendicular to the transverse axis (the $x$-axis, for our standard hyperbola). By symmetry, a hyperbola has two latus recta of equal length — one through each focus $F(c,0)$ and $F'(-c,0)$, where $c=\sqrt{a^2+b^2}$.

Why it has a clean formula. Because the latus rectum is a vertical chord at a fixed $x$-value ($x=c$), finding its endpoints is just a matter of substituting $x=c$ into the hyperbola's equation and solving for $y$. The two resulting $y$-values, being equal and opposite, hand you the chord length directly — no calculus needed, just algebra. This is exactly the same strategy used for the ellipse's latus rectum, and it is a technique worth remembering in its own right: whenever a chord is described as passing through a named point and being perpendicular to an axis, fixing that coordinate and solving the conic's equation for the other variable is usually the fastest route to its endpoints.

Why the topic matters in the syllabus. CBSE Class 11 treats the latus rectum, alongside the standard equation and the eccentricity formula, as one of three "simple properties" every student must be able to state and derive for the hyperbola. It is a small, self-contained result — a handful of algebraic steps — but it is tested often because it combines two skills at once: substituting into a conic's equation, and using the axis–focus relation $c^2=a^2+b^2$ correctly. Getting comfortable with this derivation also pays off when the same chord-through-a-focus idea reappears for the parabola later in the same chapter.

Endpoints. Substituting $x=c$ into $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ and using $c^2-a^2=b^2$ gives $y=\pm\dfrac{b^2}{a}$, so the endpoints are $\left(c,\dfrac{b^2}{a}\right)$ and $\left(c,-\dfrac{b^2}{a}\right)$. Their vertical separation is the length:

where: $L$ is the length of the latus rectum, $a$ is the semi-transverse axis (distance from centre to vertex), and $b$ is the semi-conjugate axis appearing in the standard equation.

In terms of eccentricity. Since $b^2=a^2(e^2-1)$ follows from $e=\sqrt{1+b^2/a^2}$, the same length can be written as

where: $e$ is the eccentricity of the hyperbola ($e>1$ always). This form is handy whenever a problem states $e$ directly instead of $b$.

The key contrast with the ellipse. For an ellipse, $a>b$ is enforced, so the latus rectum $\dfrac{2b^2}{a}$ is always shorter than the major axis $2a$. A hyperbola has no such restriction — $b$ can exceed $a$ freely — so its latus rectum can be shorter than, equal to, or longer than the transverse axis $2a$, and it has no fixed upper bound at all. Set $\dfrac{2b^2}{a}=2a$ and you get $b=a$, the "rectangular hyperbola" case, where the latus rectum exactly equals the transverse axis — a useful reference point to keep in mind when checking whether a given hyperbola's latus rectum looks unusually large or small.

Where the two latus recta sit. The full picture has two identical chords, one at $x=c$ through the right focus and one at $x=-c$ through the left focus, each of length $\dfrac{2b^2}{a}$, positioned symmetrically about the centre. A problem that asks for "the latus rectum" (singular) is really asking for this common length, since both chords are congruent — you never need to compute them separately.

Problem. Find the length of the latus rectum of the hyperbola $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$, and verify the result using eccentricity.

Step 1. Read off $a^2=9$ so $a=3$, and $b^2=16$.

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

Step 3. Verify via eccentricity. Compute $c^2=a^2+b^2=9+16=25$, so $c=5$ and $e=c/a=5/3$. Then $e^2-1=\dfrac{25}{9}-1=\dfrac{16}{9}$, and $2a(e^2-1)=2(3)\left(\dfrac{16}{9}\right)=\dfrac{32}{3}$, matching Step 2 exactly.

Answer: the latus rectum has length $\boldsymbol{\dfrac{32}{3}}$.

Competitive stretch

A hyperbola has transverse axis $2a=10$ and latus rectum $9.6$. Step 1. $a=5$. Step 2. From $\dfrac{2b^2}{a}=9.6$, get $2b^2=48$, so $b^2=24$. Step 3. $c^2=25+24=49\Rightarrow c=7$, so $e=c/a=7/5=1.4$. Step 4. Check: $2a(e^2-1)=2(5)(1.96-1)=10(0.96)=9.6$, matching the given latus rectum. Answer: $b^2=24$, $e=1.4$.

• Substituting $x=a$ (the vertex) instead of $x=c$ (the focus) — the latus rectum must pass through a focus, not a vertex.
• Using the ellipse's relation $c^2=a^2-b^2$ instead of the hyperbola's $c^2=a^2+b^2$, which gives the wrong endpoints entirely.
• Assuming the latus rectum must be shorter than the transverse axis $2a$ — true for the ellipse, not guaranteed for the hyperbola since $b$ has no upper bound relative to $a$.
• Using the ellipse's eccentricity shortcut $2a(1-e^2)$ instead of the hyperbola's $2a(e^2-1)$ — the sign inside the bracket flips because $e<1$ for an ellipse but $e>1$ for a hyperbola.
• Forgetting that a hyperbola has two latus recta of equal length — a question that says "find the latus recta" (plural) still just wants the one value $\dfrac{2b^2}{a}$, quoted once, since both chords are congruent by symmetry.

Ask the tutor: Want to see this latus-rectum derivation done side by side with the ellipse's, so the one-sign difference becomes completely clear?