Question

The parabolas y² = 4ax and x² = 4 by intersect orthogonally at point P(x₁, y₁) where x₁ . y₁≠ 0 provided

• A. b = a²
• B. b = a³
• C. b³ = a²
• D. None

Answer 1

Answer: (D)

None

Answer 2

Solving y² = 4ax and x² = 4by we get x = 0 or x³ = 64ab². Slope of the curves at the common points are $\frac { 2 a } { y }$ and $\frac { x } { 2 b }$ respectively. If these parabola intersect orthogonally then $\frac { 2 a } { y } \cdot \frac { x } { 2 b }$ = -1

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

⇒ x = −4a (as x ≠ 0)

⇒ −x³ = 64a³

⇒ 64ab² + 64a³ = 0

⇒ a² + b² = 0

which is not possible.