99 Percentile Qs Bank for JEE MainMathematicsParabola
Let a focal chord 12 x+5 y-27=0 of the parabola y ^2= kx intersect the parabola at the points P and P ^ . If S is the focus of this parabola, then 9 ( SP + SP ^1 )=
Options
- A27
- B108
- C16 SP.SP ^
- D4 SP . SP ^
Correct answer
D. 4 SP . SP ^
Step-by-step solution
Given that equation of parabola y^2= km then focus is ( k 4 , 0 ) which is lies on focal chord. 12 k 4 +0-27=0 k=9 , focus s ( 9 4 , 0 ) On solving the equation y^2=9 x and 12 x+5 y-27=0 We get P(1,3) and P^ ( 81 16 , -27 4 ) . aligned & S P= ( 9 4 -1 )^2+3^2 = 13 4 & S P^ = ( 81 16 - 9 4 )^2+ ( 27 4 )^2 = 117 16 aligned Now, g (S P+S P^ )= 1521 1 =4 S P . S P^ .