# 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.

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The Hyperbola