Published by:
CGP EDU Academic Team
Published on: August 13, 2026
The parabolas y 2 = 4ax and x 2 = 4by intersect orthogonally at point P(x 1 , y 1 ) where x 1 y 1 
0 then :
Text Solution
Verified by ExpertsThe correct answer is:
D
on solving y 2 = 4ax & x 2 = 4by we
get x = 0 and x 3 = 64ab 2 .
= 
, 
= 
Given curves intersect orthogonally
⇒ 
× 
= – 1
ax + by = 0
ax + 
= 0
x = – 4a (x 
0)
⇒ – x 3 = 64a 3 = – 64ab 2 ⇒ a
2 + b 2 = 0
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