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.

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$^{2} - 1$)

Notes:

(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.

Solution:

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

• Definition of Hyperbola
• Standard Equation of an Hyperbola
  
• Vertex of the Hyperbola
• Centre of the Hyperbola
• Transverse and Conjugate Axis of the Hyperbola
• Two Foci and Two Directrices of the Hyperbola
• Latus Rectum of the Hyperbola
• Position of a Point with Respect to the Hyperbola
• Conjugate Hyperbola
• Rectangular Hyperbola
• Parametric Equation of the Hyperbola
• Hyperbola Formulae
• Problems on Hyperbola

11 and 12 Grade Math 

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