Latus Rectum of the Hyperbola

We will discuss about the latus rectum of the hyperbola along with the examples.

Definition of the Latus Rectum of  the Hyperbola:

The chord of the hyperbola through its one focus and perpendicular to the transverse axis (or parallel to the directrix) is called the latus rectum of the hyperbola.

Latus Rectum of  the Hyperbola

It is a double ordinate passing through the focus. Suppose the equation of the hyperbola be \(\frac{x^{2}}{a^{2}}\) - \(\frac{y^{2}}{b^{2}}\) = 1 then, from the above figure we observe that L\(_{1}\)SL\(_{2}\) is the latus rectum and L\(_{1}\)S is called the semi-latus rectum. Again we see that M\(_{1}\)SM\(_{2}\) is also another latus rectum.

According to the diagram, the co-ordinates of the end L\(_{1}\) of the latus rectum L\(_{1}\)SL\(_{2}\) are (ae, SL\(_{1}\)). As L\(_{1}\) lies on the hyperbola \(\frac{x^{2}}{a^{2}}\) - \(\frac{y^{2}}{b^{2}}\) = 1, therefore, we get,

\(\frac{(ae)^{2}}{a^{2}}\) - \(\frac{(SL_{1})^{2}}{b^{2}}\) = 1

\(\frac{a^{2}e^{2}}{a^{2}}\) - \(\frac{(SL_{1})^{2}}{b^{2}}\) = 1     

e\(^{2}\) - \(\frac{(SL_{1})^{2}}{b^{2}}\) = 1

⇒ \(\frac{(SL_{1})^{2}}{b^{2}}\) = e\(^{2}\) - 1

⇒ SL\(_{1}\)\(^{2}\) = b\(^{2}\) . \(\frac{b^{2}}{a^{2}}\), [Since, we know that, b\(^{2}\) = a\(^{2}\)(e\(^{2} - 1\))]

⇒ SL\(_{1}\)\(^{2}\) = \(\frac{b^{4}}{a^{2}}\)       

Hence, SL\(_{1}\) = ± \(\frac{b^{2}}{a}\).

Therefore, the co-ordinates of the ends L\(_{1}\) and L\(_{2}\) are (ae, \(\frac{b^{2}}{a}\)) and (ae, - \(\frac{b^{2}}{a}\)) respectively and the length of latus rectum = L\(_{1}\)SL\(_{2}\) = 2 . SL\(_{1}\) = 2 . \(\frac{b^{2}}{a}\) = 2a(e\(^{2} - 1\))


(i) The equations of the latera recta of the hyperbola \(\frac{x^{2}}{a^{2}}\) - \(\frac{y^{2}}{b^{2}}\) = 1 are x = ± ae.

(ii) A hyperbola has two latus rectum.

Solved examples to find the length of the latus rectum of a hyperbola:

Find the length of the latus rectum and equation of the latus rectum of the hyperbola x\(^{2}\) - 4y\(^{2}\) + 2x - 16y - 19 = 0.


The given equation of the hyperbola x\(^{2}\) - 4y\(^{2}\) + 2x - 16y - 19 = 0

Now form the above equation we get,

(x\(^{2}\) + 2x + 1) - 4(y\(^{2}\) + 4y + 4) = 4

(x + 1)\(^{2}\) - 4(y + 2)\(^{2}\) = 4.

Now dividing both sides by 4

⇒ \(\frac{(x + 1)^{2}}{4}\) - (y + 2)\(^{2}\) = 1.

\(\frac{(x + 1)^{2}}{2^2} - \frac{(y + 2)^{2}}{1^{2}}\) ………………. (i)

Shifting the origin at (-1, -2) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by X and Y, we have

x = X - 1 and y = Y - 2 ………………. (ii)

Using these relations, equation (i) reduces to \(\frac{X^{2}}{2^{2}}\) - \(\frac{Y^{2}}{1^{2}}\) = 1 ………………. (iii)

This is of the form \(\frac{X^{2}}{a^{2}}\) - \(\frac{Y^{2}}{b^{2}}\) = 1, where a = 2 and b = 1.

Thus, the given equation represents a hyperbola.

Clearly, a > b. So, the given equation represents a hyperbola whose tranverse and conjugate axes are along X and Y axes respectively.

Now fine the eccentricity of the hyperbola:

We know that e = \(\sqrt{1 + \frac{b^{2}}{a^{2}}}\) = \(\sqrt{1 + \frac{1^{2}}{2^{2}}}\) = \(\sqrt{1 + \frac{1}{4}}\) = \(\frac{√5}{2}\).

Therefore, the length of the latus rectum = \(\frac{2b^{2}}{a}\) = \(\frac{2 ∙ (1)^{2}}{2}\) = \(\frac{2}{2}\) = 1.

The equations of the latus recta with respect to the new axes are X = ±ae

X = ± 2 \(\frac{√5}{2}\)

X = ± √5

Hence, the equations of the latus recta with respect to the old axes are

x = ±√5 – 1, [Putting X = ± √5 in (ii)]

i.e., x = √5 - 1 and x = -√5 – 1.

The Hyperbola

11 and 12 Grade Math 

From Latus Rectum of the Hyperbola to HOME PAGE

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Worksheet on Triangle | Homework on Triangle | Different types|Answers

    Jun 21, 24 02:19 AM

    Find the Number of Triangles
    In the worksheet on triangle we will solve 12 different types of questions. 1. Take three non - collinear points L, M, N. Join LM, MN and NL. What figure do you get? Name: (a)The side opposite to ∠L…

    Read More

  2. Worksheet on Circle |Homework on Circle |Questions on Circle |Problems

    Jun 21, 24 01:59 AM

    In worksheet on circle we will solve 10 different types of question in circle. 1. The following figure shows a circle with centre O and some line segments drawn in it. Classify the line segments as ra…

    Read More

  3. Circle Math | Parts of a Circle | Terms Related to the Circle | Symbol

    Jun 21, 24 01:30 AM

    Circle using a Compass
    In circle math the terms related to the circle are discussed here. A circle is such a closed curve whose every point is equidistant from a fixed point called its centre. The symbol of circle is O. We…

    Read More

  4. Circle | Interior and Exterior of a Circle | Radius|Problems on Circle

    Jun 21, 24 01:00 AM

    Semi-circular Region
    A circle is the set of all those point in a plane whose distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is known

    Read More

  5. Quadrilateral Worksheet |Different Types of Questions in Quadrilateral

    Jun 19, 24 09:49 AM

    In math practice test on quadrilateral worksheet we will practice different types of questions in quadrilateral. Students can practice the questions of quadrilateral worksheet before the examinations

    Read More