Question

Two of straight lines given by 3x³ +Py³ + 3x²y –3xy² = 0 are at 90⁰, if

• A. P= – $\frac { 1 } { 3 }$
• B. P = $\frac { 1 } { 3 }$
• C. P = –3
• D. P = 3

Answer 1

Answer: (C)

P = –3

Answer 2

Given equation can be written as

P– 3+ 3 $\left( \frac { y } { x } \right)$ + 3 = 0

\ Pm³ –3m² + 3m + 3 = 0 … (1)

Let m₁, m₂, m₃ are roots

\ m₁m₂m₃ = $\frac { - 3 } { \mathrm { P } }$

But two lines are perpendicular

\ m₃ = $\frac { 3 } { \mathrm { P } }$

Put in (1)

P = –3