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 x2a2 - y2b2 = 1 then, from the above figure we observe that L1SL2 is the latus rectum and L1S is called the semi-latus rectum. Again we see that M1SM2 is also another latus rectum.

According to the diagram, the co-ordinates of the end L1 of the latus rectum L1SL2 are (ae, SL1). As L1 lies on the hyperbola x2a2 - y2b2 = 1, therefore, we get,

(ae)2a2 - (SL1)2b2 = 1

a2e2a2 - (SL1)2b2 = 1     

e2 - (SL1)2b2 = 1

⇒ (SL1)2b2 = e2 - 1

⇒ SL12 = b2 . b2a2, [Since, we know that, b2 = a2(e21)]

⇒ SL12 = b4a2       

Hence, SL1 = ± b2a.

Therefore, the co-ordinates of the ends L1 and L2 are (ae, b2a) and (ae, - b2a) respectively and the length of latus rectum = L1SL2 = 2 . SL1 = 2 . b2a = 2a(e21)

Notes:

(i) The equations of the latera recta of the hyperbola x2a2 - y2b2 = 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 x2 - 4y2 + 2x - 16y - 19 = 0.

Solution:

The given equation of the hyperbola x2 - 4y2 + 2x - 16y - 19 = 0

Now form the above equation we get,

(x2 + 2x + 1) - 4(y2 + 4y + 4) = 4

(x + 1)2 - 4(y + 2)2 = 4.

Now dividing both sides by 4

⇒ (x+1)24 - (y + 2)2 = 1.

(x+1)222(y+2)212 ………………. (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 X222 - Y212 = 1 ………………. (iii)

This is of the form X2a2 - Y2b2 = 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 = 1+b2a2 = 1+1222 = 1+14 = 52.

Therefore, the length of the latus rectum = 2b2a = 2(1)22 = 22 = 1.

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

X = ± 2 52

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 

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