The One Measurement That Tells You How Wide A Curve Opens At Its Focus
Point a satellite dish at the sky and its exact shape decides how much signal it gathers. Buried in that shape is a single length that fixes how "open" the dish is right at its focus, and that length has a name: the latus rectum.
The latus rectum of a conic section is the chord that passes through a focus, runs perpendicular to the major (or transverse) axis, and has both endpoints on the curve. It gives a direct measure of how wide the conic is at the focus. A parabola has one latus rectum; an ellipse and a hyperbola each have two, one through each focus. The latus rectum is a defining feature of the conic sections family, and it is tied closely to a conic's eccentricity.
By the end you will know each conic's latus-rectum length, where its endpoints sit, and why the same $\dfrac{2b^2}{a}$ appears twice.
Latus Rectum Of A Parabola: Length $4a$
For the standard parabola opening rightward:
$$y^2 = 4ax$$
the focus is at $(a, 0)$ and the directrix is the line $x = -a$. The latus rectum is the vertical chord through the focus. To find where it meets the curve, set $x = a$:
$$y^2 = 4a(a) = 4a^2 \quad\Rightarrow\quad y = \pm 2a$$
So the endpoints are $L(a, 2a)$ and $L'(a, -2a)$, and the length is the distance between them:
$$\text{Latus rectum} = 2a - (-2a) = 4a$$
Here $a$ is the distance from the vertex to the focus. A larger $a$ opens the parabola wider, and the latus rectum $4a$ measures exactly that opening at the focus. The focus itself is the focus of a parabola, and the guiding line is the directrix of a parabola.
Latus Rectum Of An Ellipse: Length $\dfrac{2b^2}{a}$
For the standard horizontal ellipse:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \qquad a > b$$
the foci sit at $(\pm ae, 0)$, where $e$ is the eccentricity and $0 < e < 1$. Substitute the focal x-value $x = ae$ into the equation and solve for $y$; the algebra collapses to $y = \pm \dfrac{b^2}{a}$. So the endpoints of the latus rectum through the focus $(ae, 0)$ are:
$$\left(ae, \frac{b^2}{a}\right) \quad\text{and}\quad \left(ae, -\frac{b^2}{a}\right)$$
and the length is:
$$\text{Latus rectum} = \frac{2b^2}{a}$$
Here $a$ is the semi-major axis and $b$ is the semi-minor axis. Because an ellipse has two foci, it has two latus rectums, each of the same length $\dfrac{2b^2}{a}$. These pass through the two foci of the ellipse.
Latus Rectum Of A Hyperbola: length $\dfrac{2b^2}{a}$
For the standard horizontal hyperbola:
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
the foci sit at $(\pm ae, 0)$, where the eccentricity $e > 1$. Substituting $x = ae$ and solving for $y$ again gives $y = \pm \dfrac{b^2}{a}$, so the latus-rectum endpoints through $(ae, 0)$ are:
$$\left(ae, \frac{b^2}{a}\right) \quad\text{and}\quad \left(ae, -\frac{b^2}{a}\right)$$
and the length is the same expression as the ellipse:
$$\text{Latus rectum} = \frac{2b^2}{a}$$
The two conics share the formula because both use $b^2$ tied to the focal geometry; the difference lives in the equation's sign and in the range of $e$, not in the latus-rectum length. The two chords pass through the two foci of the hyperbola.
Why does the same $\dfrac{2b^2}{a}$ serve both the ellipse and the hyperbola? In each case the semi-latus rectum equals $\dfrac{b^2}{a}$, and doubling it gives the full chord. The ellipse and hyperbola both build $b^2$ from the same focal relationship, so the width-at-focus formula matches even though the curves look nothing alike.
Examples of Latus Rectum
Example 1
Find the length and endpoints of the latus rectum of the parabola $y^2 = 12x$.
Compare with $y^2 = 4ax$:
$$4a = 12 \quad\Rightarrow\quad a = 3$$
Length of latus rectum:
$$4a = 12$$
Endpoints, using $(a, \pm 2a)$:
$$(3, 6) \quad\text{and}\quad (3, -6)$$
The latus rectum has length $12$, with endpoints $(3, 6)$ and $(3, -6)$.
Example 2
A student reports the latus rectum of $y^2 = 12x$ as $a = 3$ instead of $4a$. Spot the error.
A natural first move is to read off $a = 3$ and stop, treating $a$ as the answer. But $a$ is only the focus-to-vertex distance, not the chord length, and calling it the latus rectum confuses a coordinate with a length.
The latus-rectum length is $4a$, so:
$$4a = 4(3) = 12$$
The length is $12$. Always finish with $4a$; the bare $a$ is just an intermediate value.
Example 3
Find the length of the latus rectum of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.
Read off $a^2 = 25$ and $b^2 = 9$, so $a = 5$ and $b = 3$. Since $a > b$, the major axis is horizontal, and the formula applies directly:
$$\text{Latus rectum} = \frac{2b^2}{a} = \frac{2(9)}{5} = \frac{18}{5} = 3.6$$
The latus rectum has length $3.6$.
Example 4
Find the length of the latus rectum of the hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$.
Read off $a^2 = 16$ and $b^2 = 9$, so $a = 4$ and $b = 3$:
$$\text{Latus rectum} = \frac{2b^2}{a} = \frac{2(9)}{4} = \frac{18}{4} = 4.5$$
The latus rectum has length $4.5$.
Example 5
An ellipse has latus rectum $\dfrac{2b^2}{a} = 8$ and semi-major axis $a = 4$. Find $b$.
Substitute the known values:
$$\frac{2b^2}{4} = 8$$
$$\frac{b^2}{2} = 8 \quad\Rightarrow\quad b^2 = 16 \quad\Rightarrow\quad b = 4$$
So $b = 4$. Since $a = b = 4$ here, the ellipse is in fact a circle, a useful check that the widest possible ellipse (a circle) has the largest latus rectum for its size.
Example 6
A parabolic satellite dish is modelled by $y^2 = 4ax$ and must have a latus rectum of $2$ metres so its receiver spans the right width at the focus. Find $a$ and the focus position.
The latus rectum is $4a$:
$$4a = 2 \quad\Rightarrow\quad a = 0.5 \text{ m}$$
The focus sits at $(a, 0) = (0.5, 0)$, so the receiver is placed $0.5$ m from the vertex along the axis. The latus rectum of $2$ m tells the engineer how wide the beam is where the receiver sits, which is why the measurement matters in dish design.
Where The Latus Rectum Earns Its Keep: Width At The Focus
The latus rectum matters because it converts an abstract focus into a concrete size: how wide the curve is right where the action happens.
Optics and antennas. A parabolic mirror or dish focuses signal at its focus; the latus rectum $4a$ fixes how broad the beam is there, which sets receiver size and gain.
Orbits. In planetary motion, the semi-latus rectum $\dfrac{b^2}{a}$ is the standard parameter for an orbit's shape, appearing directly in the polar equation of a conic used across celestial mechanics.
Why it is the natural width. The destination is a focus-anchored size. The vertex or centre gives one landmark, but the focus is where light, gravity, and signal concentrate, so the chord through the focus is the width that physics actually uses.
The semi-latus rectum is the shape parameter in the conic-section orbit equation, which is why astronomers describe orbits by this length rather than by the axes alone.
Mistakes to watch for
Mistake 1: Reporting $a$ instead of $4a$ for a parabola
Where it slips in: Finding the latus rectum right after solving $4a = \text{coefficient}$.
Don't do this: State the latus rectum as $a$, the focus-to-vertex distance.
The correct way: The latus-rectum length is $4a$; $a$ is only an intermediate. For $y^2 = 12x$, $a = 3$ but the latus rectum is $12$. Finish the calculation rather than stopping at $a$.
Mistake 2: Swapping $a$ and $b$ in $\dfrac{2b^2}{a}$
Where it slips in: Applying the ellipse or hyperbola formula when the equation lists $a^2$ and $b^2$ in either order.
Don't do this: Write $\dfrac{2a^2}{b}$ by grabbing the larger number for the top.
The correct way: In the standard horizontal form, $a$ is tied to the $x^2$ term and $b$ to the $y^2$ term; the numerator is $2b^2$ and the denominator is $a$. Identify which is which from the equation before substituting. The memoriser who stores "$2b^2/a$" without checking which letter goes where flips them on the first vertical ellipse.
Mistake 3: Forgetting a conic can have two latus rectums
Where it slips in: Describing the full geometry of an ellipse or hyperbola.
Don't do this: Draw a single latus rectum and treat the ellipse like a parabola.
The correct way: A parabola has one focus and one latus rectum; an ellipse and a hyperbola each have two foci, so each has two latus rectums of equal length. Account for both when a problem asks for the whole figure.
Key Takeaways
The latus rectum is the focal chord perpendicular to the main axis, ending on the curve.
For a parabola $y^2 = 4ax$, its length is $4a$, with endpoints $(a, \pm 2a)$.
For an ellipse and a hyperbola, its length is $\dfrac{2b^2}{a}$, with endpoints $\left(ae, \pm\dfrac{b^2}{a}\right)$.
A parabola has one latus rectum; the ellipse and hyperbola each have two.
The latus rectum measures a conic's width at the focus, which is why it appears in optics and orbits.
A practical next step
Practice these problems to solidify your understanding. For each conic, first match the equation to its standard form, then apply the right latus-rectum formula.
Find the latus rectum of the parabola $y^2 = 20x$. (Answer to Question 1: $4a = 20$.)
Find the latus rectum of the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$. (Answer to Question 2: $\dfrac{2(16)}{6} = \dfrac{32}{6} \approx 5.33$.)
To work through conic sections with a teacher, explore Bhanzu's geometry tutor, our high school math tutor sessions, or math classes online. To see a trainer derive each conic's latus rectum live, you can book a free demo class.
Read More
Parabola - the curve whose latus rectum is $4a$.
Ellipse - where the $\dfrac{2b^2}{a}$ latus rectum first appears.
Hyperbola - the two-branch conic that shares the $\dfrac{2b^2}{a}$ formula.
Eccentricity of ellipse - how the ellipse's shape parameter relates to its latus rectum.
Area of ellipse - the $\pi a b$ area built from the same $a$ and $b$ as the latus rectum.
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