Question

The parabolas y² = 4ax and x² = 4by intersect orthogonally at point P(x₁, y₁) where x₁ y₁$\neq$0 then :

• A. b = a²
• B. b = a³
• C. b³ = a²
• D. a² + b² = 0

Answer 1

Answer: (D)

a² + b² = 0

Answer 2

on solving y² = 4ax & x² = 4by we

get x = 0 and x³ = 64ab².

= , =

Given curves intersect orthogonally

Ž × $\frac { x } { 2 b }$ = – 1

ax + by = 0

ax + $\frac { x ^ { 2 } } { 4 }$ = 0

x = – 4a (x $\neq$ 0)

Ž – x³ = 64a³ = – 64ab²

Ž a² + b² = 0