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.

Latus rectum of the hyperbola

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

From the latus rectum of the hyperbola to the HOMEPAGE


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