PROVE THAT THE LENGTH OF LATUS RECTUM OF THE HYPERBOLA

About "Prove that the Length of the Latus Rectum of the Hyperbola"

Prove that the Length of the Latus Rectum of the Hyperbola :

Here we are going to see how to prove the length of latus rectum of the hyperbola as 2b²/a

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)²  =  b²(e² - 1)  -----(1)

b²  =  a² (e² - 1)

(e² - 1)  =  b²/a²

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

(SL)²  =  b²(b²/a²) 

(SL)²  =  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.

After having gone through the stuff given above, we hope that the students would have understood, "Prove that the Length of the Latus Rectum of the Hyperbola".

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