PROVE THAT THE LENGTH OF LATUS RECTUM OF THE HYPERBOLA

Question  1 :

Prove that the length of the latus rectum of the hyperbola (x2/a2) - (y2/b2)  =  1 is 2b2/a.

Solution :

In the picture given above LSL' is the latus rectum and LS is called semi latus rectum TS'T' is also a latus rectum.

The coordinates of L are (ae, SL)

As L lies on the hyperbola.

(x2/a2) - (y2/b2)  =  1

The coordinate will satisfy the equation of the hyperbola

((ae)2/a2) - ((SL)2/b2)  =  1

(a2e2/a2) - ((SL)2/b2)  =  1

e2 - 1  =  ((SL)2/b2)

(SL) =  b2(e2 - 1)  -----(1)

b =  a(e2 - 1)

(e2 - 1)  =  b2/a2

By applying the value of (e2 - 1) in (1), we get 

(SL) =  b2(b2/a2

(SL) =  b4/a2 

SL  =  b2/a (length of semi latus rectum)

SL + SL'  =  2b2/a

Hence proved.

Question 2 :

Show that the absolute value of difference of the focal distances of any point P on the hyperbola is the length of its transverse axis

Solution :

Let p(x,y) be any point on the hyperbola

(x2/a2) - (y2/b2)  =  1.a

Let MPM' be the perpendicular through P on directrices ZK and Z'K'. Now by definition we get,

SP = e  PM

SP = e ∙ NK

SP = e (CN - CK)

SP = e(X - (a/e))

SP = eX - a

and

S'P = e  PM'

⇒ S'P = e  (NK')

⇒ S'P = e (CK' + CN)

⇒ S'P = e(X + (a/e))

S'P = eX + a

Therefore,

S'P - SP = (a + ex) - (ex - a)

  =  a + ex - ex + a

  =  2a

= length of transverse axis.

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