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

Definition of the latus rectum of the hyperbola:

The tendon of the hyperbola by its one focal point and perpendicular to the transverse axis (or parallel to the guideline) is called the rectum Hyperbole.

It is a double ordinate that goes through focus. Assume the equation of the Be hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 then from the above figure we note that that L (_ {1} ) SL (_ {2} ) is the latus rectum and L (_ {1} ) S is called the semilatus rectum. Again we see that M (_ {1} ) SM (_ {2} ) is also another latus rectum.

According to the diagram, the coordinates of the end are L. (_ {1} ) of the rectum L (_ {1} ) SL (_ {2} ) are (ae, SL(_{1})). As L (_ {1} ) lies on the hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1, hence 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} ).

Hence the coordinates of the ends are L. (_ {1} ) and L (_ {2} ) are (ae, ( frac {b ^ {2}} {a} )) and (ae, – ( frac {b ^ {2}} {a} )) or the length of the rectum = L (_ {1} ) SL (_ {2} ) = 2. SL (_ {1} ) = 2. ( frac {b ^ {2}} {a} ) = 2a (e (^ {2} – 1 ))

**Remarks: **

(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 determine the length of the latus rectum of a hyperbola:

Find the length of the latus rectum and the equation of the latus rectum des Hyperbola x (^ {2} ) – 4y (^ {2} ) + 2x – 16y – 19 = 0.

**Solution:**

The given equation of Hyperbola x (^ {2} ) – 4y (^ {2} ) + 2x – 16y – 19 = 0

Now form the above equation that we get,

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

⇒ (x + 1) (^ {2} ) – 4 (y + 2) (^ {2} ) = 4.

Now divide 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)

Moving the origin at (-1, -2) without rotating the coordinate axes and denoting the new coordinates in relation to the new axes with X and Y, we have

x = X – 1 and y = Y – 2 ………………. (ii)

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

It’s the shape ( frac {X ^ {2}} {a ^ {2}} ) – ( frac {Y ^ {2}} {b ^ {2}} ) = 1, where a = 2 and b = 1.

Thus the given equation represents a Hyperbole.

Obviously a> b. So the given equation represents ahyperbole whose transverse and conjugate axes lie along the X and Y axes, respectively.

Well the eccentricity of the Hyperbole:

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} ) = ( frak {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

The equations of the latus recta with respect to the ancient axes are thus

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

ie x = 5 – 1 and x = –√5 – 1.

**● ****The ****hyperbole**

**11th and 12th grade math degree**

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