Latus Rectum: Parabola, Ellipse, and Hyperbola Formulas

TL;DR

The latus rectum is the focal chord of a conic drawn perpendicular to its main axis, ending on the curve; its length is 4a for a parabola (y^2=4ax) and (\frac{2b^2}{a}) for both an ellipse and a hyperbola. This article defines the latus rectum, gives each conic's formula and endpoint coordinates, explains why it measures a conic's width at the focus, and works through examples.

The One Measurement That Tells You How Wide A Curve Opens At Its Focus

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.

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\Rightarrow 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 (\frac{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 \frac{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 (\frac{2b^2}{a}). These pass through the two foci of the ellipse.

Latus Rectum Of A Hyperbola: Length (\frac{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 \frac{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.

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\Rightarrow a=3]

Length of latus rectum:

[4a=12]

Endpoints, using ((a,\pm 2a)):

[(3,6)\text{ and } (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). 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 12, so:

[4a=12 \Rightarrow a=3]

Example 3

Find the length of the latus rectum of the ellipse (\frac{x^2}{25} + \frac{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]

Example 4

Find the length of the latus rectum of the hyperbola (\frac{x^2}{16} - \frac{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]

Example 5

An ellipse has latus rectum (\frac{2b^2}{a} = 8) and semi-major axis (a=4). Find (b).

Substitute the known values:

[\frac{2b^2}{4} = 8\Rightarrow b^2=16 \Rightarrow b=4]

Example 6

A parabolic satellite dish is modelled by (y^2=4ax) and must have a latus rectum of 2 metres. Find (a) and the focus position.

The latus rectum is (4a):

[4a=2\Rightarrow a=0.5 \text{ m}]

The focus sits at ((a,0)=(0.5,0)) which is 0.5 m from the vertex along the axis.

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 fixes how broad the beam is there.

• Orbits. In planetary motion, the semi-latus rectum (\frac{b^2}{a}) is the standard parameter for an orbit's shape.

• Why it is the natural width. The destination is a focus-anchored size.

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 ((ae,\pm \dfrac{b^2}{a})).

• A parabola has one latus rectum; the ellipse and hyperbola each have two.

• The latus rectum measures a conic's width at the focus.