What Is the Latus Rectum of an Ellipse in Geometry?
The latus rectum of an ellipse is a focal chord perpendicular to the major axis. Breaking that down:
Focal chord: it passes through a focus of the ellipse.
Perpendicular to the major axis: it stands at right angles to the long axis.
Endpoints on the curve: both ends touch the ellipse.
For $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with semi-major axis $a$ and semi-minor axis $b$ (so $a > b$), the foci sit at $(\pm ae, 0)$ where $e$ is the eccentricity. The half-length of the latus rectum, called the semi-latus rectum, is $\frac{b^2}{a}$, so the full chord is $\frac{2b^2}{a}$.
What Is the Widest Chord You Can Draw Through an Ellipse's Focus?
An ellipse hides one chord at its focus whose length never changes for that shape.
The latus rectum of an ellipse is the chord that passes through a focus and runs perpendicular to the major axis, with both ends on the ellipse. For the standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$, the length of each latus rectum is $\frac{2b^2}{a}$. Because an ellipse has two foci, it has two latus recta - one through each focus - and they are equal in length.
The Latin name means "straight side," and the chord is exactly that: a fixed measuring rod planted at the focus that captures how "wide" the ellipse is at its focal points. It is one of the standard quantities that describe an ellipse, alongside its axes and its foci.
What Is the Length of the Latus Rectum of an Ellipse, and How Is It Derived?
The formula is worth deriving once so it stops being something to memorise cold. The length is $\frac{2b^2}{a}$, and here is where it comes from.
The right focus sits at $x = ae = c$, where $c = \sqrt{a^2 - b^2}$. Substitute $x = c$ into the ellipse equation to find the $y$-values of the chord's endpoints:
$$\frac{c^2}{a^2} + \frac{y^2}{b^2} = 1$$
Solve for $y^2$:
$$\frac{y^2}{b^2} = 1 - \frac{c^2}{a^2} = \frac{a^2 - c^2}{a^2} = \frac{b^2}{a^2}$$
since $a^2 - c^2 = b^2$. Therefore:
$$y^2 = \frac{b^4}{a^2}, \qquad y = \pm \frac{b^2}{a}$$
The two endpoints are $\left(c, \frac{b^2}{a}\right)$ and $\left(c, -\frac{b^2}{a}\right)$. The distance between them — the full latus rectum — is:
$$\text{Length} = \frac{b^2}{a} - \left(-\frac{b^2}{a}\right) = \frac{2b^2}{a}$$
Variable glossary: $a$ is the semi-major axis, $b$ is the semi-minor axis, $c = ae$ is the focal distance from the centre, and $e$ is the eccentricity. At Bhanzu, our trainers teach this substitution derivation so students can rebuild the formula on any exam instead of hoping they remember it.
How Does the Latus Rectum Relate to Eccentricity and the Other Conics?
The latus rectum also has a clean form in terms of eccentricity. Since $b^2 = a^2(1 - e^2)$:
$$\frac{2b^2}{a} = \frac{2a^2(1 - e^2)}{a} = 2a(1 - e^2)$$
So a nearly circular ellipse (small $e$) has a latus rectum close to $2a$, while a highly elongated ellipse (large $e$) has a short one. The same eccentricity that sets the latus rectum also fixes the directrix of the ellipse, the lines $x = \pm \tfrac{a}{e}$ that pair with each focus. The concept generalises across all conic sections: the parabola and hyperbola each have their own latus rectum, covered together in the latus rectum of parabola, ellipse, and hyperbola overview. For the parabola specifically, the latus rectum has length $4p$ and behaves differently, since a parabola has only one focus.
What Are the Properties of the Latus Rectum of an Ellipse?
Gathering the key facts in one place makes the latus rectum easy to reason about:
There are two equal latus recta. Because an ellipse has two foci, it carries two latus recta, one through each focus, and both share the same length $\frac{2b^2}{a}$.
Each is perpendicular to the major axis. A latus rectum stands at right angles to the major axis, which makes it parallel to the minor axis and to the directrix.
Its endpoints are collinear with the focus. For the right focus, the two endpoints $\left(ae, \frac{b^2}{a}\right)$ and $\left(ae, -\frac{b^2}{a}\right)$ and the focus $(ae, 0)$ all lie on one vertical line.
It is always shorter than the major axis. Since $\frac{2b^2}{a} = 2a(1 - e^2)$ and $0 < e < 1$ for every ellipse, the latus rectum is less than $2a$; it grows toward $2a$ as the ellipse rounds into a circle and shrinks as the ellipse elongates.
The semi-latus rectum fixes the shape at the focus. The half-length $\frac{b^2}{a}$ is the single parameter that captures how wide the ellipse is at its focal points, which is why it is the value that appears in the orbit equation.
Examples of the Latus Rectum of Ellipse
Example 1
Find the length of the latus rectum of $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Here $a^2 = 25$ so $a = 5$, and $b^2 = 16$ so $b = 4$. Since $a > b$:
$$\text{Length} = \frac{2b^2}{a} = \frac{2 \times 16}{5} = \frac{32}{5} = 6.4$$
Final answer: $\frac{32}{5} = 6.4$ units.
Example 2
Find the latus rectum of $\frac{x^2}{9} + \frac{y^2}{16} = 1$ — watch the axis trap.
Wrong path. A student reads off $a^2 = 9$, $b^2 = 16$ and writes $\frac{2b^2}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$.
Why it breaks. In this ellipse the larger denominator is under $y^2$, so the major axis is vertical. The semi-major axis is the bigger value: $a = 4$, and the semi-minor is $b = 3$. The student swapped them.
Correct. With $a = 4$ and $b = 3$:
$$\text{Length} = \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{18}{4} = 4.5$$
Final answer: $4.5$ units. The value $a$ must always be the semi-major axis - the largest denominator's root.
Example 3
Find the endpoints of the latus rectum through the right focus of $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
Here $a = 5$, $b = 3$, and $c = \sqrt{25 - 9} = \sqrt{16} = 4$. The semi-latus rectum is $\frac{b^2}{a} = \frac{9}{5} = 1.8$. The right focus is at $(4, 0)$.
Final answer: endpoints $(4, 1.8)$ and $(4, -1.8)$.
Example 4
An ellipse has semi-major axis $a = 6$ and eccentricity $e = \tfrac{1}{2}$. Find the latus rectum.
Use the eccentricity form: $\frac{2b^2}{a} = 2a(1 - e^2) = 2 \times 6 \times \left(1 - \tfrac{1}{4}\right) = 12 \times \tfrac{3}{4} = 9$.
Final answer: $9$ units.
Example 5
The latus rectum of an ellipse is 8 and its semi-major axis is 9. Find $b$.
From $\frac{2b^2}{a} = 8$ with $a = 9$: $2b^2 = 8 \times 9 = 72$, so $b^2 = 36$ and $b = 6$.
Final answer: $b = 6$.
Example 6
Find the latus rectum of $4x^2 + 9y^2 = 36$.
First put it in standard form by dividing by 36: $\frac{x^2}{9} + \frac{y^2}{4} = 1$. So $a^2 = 9$, $a = 3$; $b^2 = 4$, $b = 2$. Since $a > b$:
$$\text{Length} = \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3} \approx 2.67$$
Final answer: $\frac{8}{3}$ units.
Why Does the Latus Rectum Matter Beyond the Classroom?
The latus rectum is not just an exam quantity - it is the natural measure of a conic's shape at its focus, and that focus is where the physics happens.
Orbital mechanics. Planets and comets travel on elliptical orbits with the Sun at a focus. The semi-latus rectum $\frac{b^2}{a}$ is the standard parameter in the orbit equation, describing the orbit's width at the focus.
Optics and antennas. Elliptical and parabolic reflectors focus signals at a focus; the latus rectum sets the aperture width at that point.
Shape comparison. Two ellipses with the same major axis but different latus recta have different "roundness" — the chord captures at a glance how eccentric the curve is.
The orbital version makes the stakes real. Johannes Kepler's discovery that planets move on ellipses with the Sun at one focus - his first law of planetary motion - put the ellipse's focus, and the focal chord that measures it, at the centre of modern astronomy.
Where Do Students Trip Up on the Latus Rectum of an Ellipse?
Mistake 1: Swapping the semi-major and semi-minor axes
Where it slips in: When the larger denominator sits under $y^2$ (a vertical ellipse), students still take $a^2$ from under $x^2$.
Don't do this: Assuming $a$ always comes from the $x^2$ term.
The correct way: $a$ is the semi-major axis - the square root of the larger denominator, whichever term it sits under. The first-instinct error here is reading positions rather than sizes, and it flips the whole formula.
Mistake 2: Confusing the ellipse formula with the parabola's
Where it slips in: Reaching for $4a$ (the parabola's latus rectum) on an ellipse problem.
Don't do this: Using $4a$ or $4p$ for an ellipse.
The correct way: The ellipse latus rectum is $\frac{2b^2}{a}$; the parabola's is $4p$. They come from different curves with different numbers of foci - keep the two formulas separate.
Mistake 3: Forgetting there are two latus recta
Where it slips in: Answering "one" when asked how many latus recta an ellipse has, or finding endpoints for only one focus.
Don't do this: Treating the ellipse like a parabola with a single focus.
The correct way: An ellipse has two foci, so it has two latus recta - one through each focus - and they are equal in length. The second-guesser often stops at one; the shape is symmetric, so both exist.
Conclusion
The latus rectum of an ellipse is the focal chord perpendicular to the major axis, with length $\frac{2b^2}{a}$.
Its endpoints are $\left(\pm ae, \pm \frac{b^2}{a}\right)$, and the semi-latus rectum is $\frac{b^2}{a}$.
In terms of eccentricity, the length equals $2a(1 - e^2)$.
Always take $a$ as the semi-major axis - the root of the larger denominator.
An ellipse has two equal latus recta, one through each focus.
To go further with a teacher, explore Bhanzu's geometry tutor or high school math tutor, or browse math classes online for conic-section practice.
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Practice These to Solidify Your Understanding
Work through these three problems in order:
Find the latus rectum of $\frac{x^2}{36} + \frac{y^2}{20} = 1$.
Find the endpoints of the latus rectum through the left focus of $\frac{x^2}{16} + \frac{y^2}{7} = 1$.
An ellipse has $b = 5$ and latus rectum 10. Find the semi-major axis $a$.
Want a live Bhanzu trainer to work through more ellipse problems with you? Book a free demo class.
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