Latus Rectum Of Ellipse - Definition, Formula, Properties and Examples

What is the Latus rectum of the ellipse?

Double ordinate passing through focus is called the latus rectum. The Latus rectum of an ellipse is a straight line passing through the foci of the ellipse and perpendicular to the major axis of the ellipse. The Latus rectum is the focal chord, which is parallel to the directrix of the ellipse. The ellipse has two foci and hence it has two latus rectums.

End Points of Latus rectum

$\mathrm{L}=\left(\mathrm{ae}, \frac{\mathrm{b}^2}{\mathrm{a}}\right)$ and $\mathrm{L}^{\prime}=\left(\mathrm{ae},-\frac{\mathrm{b}^2}{\mathrm{a}}\right)$

Length of Latus rectum

The Distance between the length of the endpoints of the latus rectum is called the Length of the Latus rectum.

The length of the latus rectum is calculated by 2b² / a

Derivation of Length of Latus Rectum

Let Latus rectum $\mathrm{LL}^{\prime}=2 \alpha$
$\mathrm{S}(\mathrm{ae}, 0)$ is focus, then $\mathrm{LS}=\mathrm{SL}^{\prime}=\alpha$
Coordinates of L and $\mathrm{L}^{\prime}$ become ( $\mathrm{ae}, \alpha$ ) and ( $\mathrm{ae},-\alpha$ ) respectively
Equation of ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Put $x=a e, y=\alpha$ we get, $\frac{(\mathrm{ae})^2}{\mathrm{a}^2}+\frac{\alpha^2}{\mathrm{~b}^2}=1 \Rightarrow \alpha^2=\mathrm{b}^2\left(1-\mathrm{e}^2\right)$
$\alpha^2=\mathrm{b}^2\left(\frac{\mathrm{b}^2}{\mathrm{a}^2}\right) \quad\left[\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\right]$
$\alpha=\frac{\mathrm{b}^2}{\mathrm{a}}$
$\Rightarrow 2 \alpha=\mathrm{LL}^{\prime}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}$

Properties Of Latus Rectum

The important properties of the latus rectum of the ellipse are as follows.

• The latus rectum is perpendicular to the major axis of the ellipse.
• The latus rectum of the ellipse passes through the focus of the ellipse.
• There are two latus rectums for an ellipse.
• Each latus rectum cuts the ellipse at two distinct points.
• The latus rectum is parallel to the directrix of the ellipse.

Terms Related to Latus Rectum of Ellipse

The following terms are related to the latus rectum of the ellipse:

1) Foci of Ellipse: The focus of the ellipse lies on the major axis of the ellipse. The ellipse has two foci and their coordinates is (+ae, 0), and (-ae, 0). The midpoint of the foci of the ellipse is the center of the ellipse.

2) Focal Chord: The line passing through the focus of the ellipse is the focal chord of the ellipse. The ellipse has an infinite number of focal chords passing through the focus.

3) Directrix: A directrix is a line that is drawn outside the ellipse and is perpendicular to the major axis of the ellipse.

4) Vertex of Ellipse: A vertex of an ellipse is the point of intersection of the ellipse with its axis of symmetry. The ellipse intersects its axis of symmetry at two distinct points, and hence an ellipse has two vertices.

5) Major Axis of Ellipse: The major axis of the ellipse is a line that cuts the ellipse into two equal halves. The major axis is a line passing through the foci and the center of the ellipse.

6) Minor Axis of Ellipse: The minor axis of the ellipse is the axis that is perpendicular to its major axis. The minor axis also passes through the center of the ellipse.

Focal Distance of a Point

The sum of the focal distance of any point on the ellipse is equal to the major axis.

Let P(x, y) be any point on the ellipse.

Here,

$
\begin{aligned}
& S P=e P M=e\left(\frac{a}{e}-x\right)=a-e x \\
& S^{\prime} P=e P M^{\prime}=e\left(\frac{a}{e}+x\right)=a+e x
\end{aligned}
$

Now, SP + S’P = a – ex + a + ex = 2a = AA' = constant.

Thus the sum of the focal distances of a point on the ellipse is constant.

Solved Examples Based on the Length of the Latus Rectum

Example 1: Let the eccentricity of an ellipse $\frac{\mathrm{a}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ is reciprocal to that of the hyperbola $2 \mathrm{x}^2-2 \mathrm{y}^2=1$. If the ellipse intersects the hyperbola at right angles, then the square of the length of the latus-rectum of the ellipse is $\qquad$ : [JEE MAINS 2023]

Solution

$
\begin{aligned}
& E: \frac{x^4}{a^2}+\frac{y^2}{b^2}=1 \rightarrow e \\
& H: x^2-y^2=\frac{1}{2} \Rightarrow e^{\prime}=\sqrt{2} \\
& e=\frac{1}{\sqrt{2}} \\
& \because e^2=\frac{1}{2} \\
& 1-\frac{b^2}{a^2}=\frac{1}{2} \Rightarrow \frac{b^2}{a^2}=\frac{1}{2} \\
& a^2=2 b^2
\end{aligned}
$

$\mathrm{E} \& \mathrm{H}$ are at a right angle they are confocal Focus of Hyperbola = focus of ellipse

$
\begin{aligned}
& \left( \pm \frac{1}{\sqrt{2}} \cdot \sqrt{2}, 0\right)=\left( \pm \frac{a}{\sqrt{2}}, 0\right) \\
& a=\sqrt{2} \\
& \because a^2=2 b^2 \Rightarrow b^2=1
\end{aligned}
$
Length of $\mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2(1)}{\sqrt{2}}$

$
=\sqrt{2}
$
Square of $\mathrm{LR}=2$
Hence, the answer is the 2 .

Example 2: If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12 , then the length of its latus rectum is :
[JEE MAINS 2019]
Solution: Given

$
\begin{aligned}
& \quad 2 a e=6 \\
& \therefore a e=3
\end{aligned}
$
Also,

$
\begin{aligned}
& \quad \frac{2 a}{e}=126 \\
& \therefore \frac{a}{e}=63
\end{aligned}
$

from (1) and (2)

$
e=\frac{1}{\sqrt{2}}, a=3 \sqrt{2}
$

since, $b^2=a^2\left(1-e^2\right)$
Substitute the values of 'e' and 'a' in the above equation.

$
\begin{aligned}
& \Rightarrow b^2=9 \\
\therefore b & = \pm 3
\end{aligned}
$

length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 9}{3 \sqrt{2}}=3 \sqrt{2}$
Hence, the answer is $3 \sqrt{2}$

Example 3: In an ellipse, with the center at the origin, if the difference of the lengths of the major axis and the minor axis is 10 and one of the foci is at $(0,5 \sqrt{3})$ then the length of its latus rectum is :
[JEE MAINS 2019]
Solutions: Given,
focus is at $(0,5 \sqrt{3})$
given the difference of the major axis-minor axis $=10$

$
\begin{aligned}
& b-a=5 \\
& b e=5 \sqrt{3} \\
& a^2=b^2\left(1-e^2\right)=b^2-(b e)^2 \\
& b=10, a=5
\end{aligned}
$
Length of $\mathrm{LR}=\frac{2 a^2}{b}=5$
Hence, the answer is 5

Example 4: Let the length of the latus rectum of an ellipse with its major axis along the $x$-axis and center at the origin be 8 . If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?
[JEE MAINS 2019]
Solution: Given, the length of the Latus rectum, $\frac{2 b^2}{a}=8$ $\qquad$

$
\begin{aligned}
& 2 a e=2 b \\
& \Rightarrow e=\frac{b}{a} \\
& \Rightarrow e^2=\frac{b^2}{a^2} \\
& \Rightarrow e^2=1-e^2 \\
& \Rightarrow e=\frac{1}{\sqrt{2}}
\end{aligned}
$
Using (ii)

$
\frac{b}{a}=\frac{1}{\sqrt{2}}
$
Using (i)
$b \cdot \frac{b}{a}=4$

$
\Rightarrow b=4 \sqrt{2} \text { and } a=8
$
So, the equation of the ellipse is

$
\frac{x^2}{64}+\frac{y^2}{32}=1
$
Hence, the answer is $(4 \sqrt{3}, 2 \sqrt{2})$

Example 5: If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is $3 / 2$ units, then its eccentricity is
[JEE MAINS 2018]
Solution: Given the length of $L R=4$

$
\frac{2 b^2}{a}=4 \Rightarrow b^2=2 a
$
And the distance between the focus and the nearest vertex

$
a-a e=3 / 2 \Rightarrow a(1-e)=3 / 2
$
Also, for an ellipse

$
\begin{aligned}
& b^2=a^2\left(1-e^2\right) \\
& 2 a=a^2\left(1-e^2\right) \\
& 2=a(1-e)(1+e) \\
& \frac{2}{1+e}=a(1-e)
\end{aligned}
$
$
\frac{2}{1+e}=3 / 2 \Rightarrow e=1 / 3
$
Hence, the answer is $1 / 3$