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.
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