JEE Main202127 Aug 2021Morning ShiftMathematicsStraight LinesActual
Let A be a fixed point ( 0 , 6 ) and B be a moving point ( 2 t , 0 ) . Let M be the mid-point of A B and the perpendicular bisector of A B meets the y - axis at C . The locus of the mid-point P of MC is
Options
- A3 x 2 + 2 y - 6 = 0
- B2 x 2 - 3 y + 9 = 0
- C3 x 2 - 2 y - 6 = 0
- D2 x 2 + 3 y - 9 = 0
Correct answer
D. 2 x 2 + 3 y - 9 = 0
Step-by-step solution
Mid point M = t , 3 m AB = - 3 t Perpendicular bisector of AB is y - 3 = t 3 x - t So, C = 0 ,   3 - t 2 3 Let P be h ,   k h = t 2 ;   k = 3 - t 2 6 ⇒ k = 3 - 4 h 2 6 ⇒ 2 x 2 + 3 y - 9 = 0