Published by:
CGP EDU Academic Team
Published on: August 13, 2026
For the hyperbola 
– 
= 1. The normal at P meets the transverse axis AA ′ in G and the conjugate axis BB ′ in g and CF be perpendicular to the normal from the centre.
(i) PF . PG = K CB 2 , then k =
A
2 CA 2 
B
1 CF 2 
C
CB 2 2 
D
4 (ii) PF . Pg equal to – CA . CB (iii) Locus of middle point of G and g is a hyperbola of eccentricity – 
Text Solution
Verified by ExpertsThe correct answer is:
A
Ans.
(i)
Sol. 1
(ii)
Sol. CA 2
(iii)
Sol. 
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