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Question : 58 of 160
Marks:
+1,
-0
Solution:
Let focal chord be latus rectum then it is given that it subtends
90° at its cantre.
Here, equation of hyperbola is
−=1 ...(i)
So eccentricity,
e=√1+ End points of latus rectum,
L=(ae,) and
L′=(ae,) Centre
(0,0) ∵<LCL′=90° In
ΔLCL′ by Pythagoras theorem
(LC)2+(L′C)2=(LL′)2 (ae−0)2+(−0)2+(ae−0)2 +(−−0)2=()2 ⇒2(a2e2+)= ⇒b4=a4e2 ⇒(e2−1)2=e2[∵b2=a2(e2−1)]
⇒e2±e−1=0 ⇒e=±= ∵e>1 for hyperbola
⇒e=, ⇒e=
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