JEE Main202528 Jan 2025Evening ShiftMathematicsHyperbolaActual
If A and B are the points of intersection of the circle x^2+y^2-8 x=0 and the hyperbola x^2 9 - y^2 4 =1 and a point P moves on the line 2 x-3 y+4=0 , then the centroid of PAB lies on the line :
Options
- Ax+9 y=36
- B4 x-9 y=12
- C6 x-9 y=20
- D9 x-9 y=32
Correct answer
C. 6 x-9 y=20
Step-by-step solution
C: x^2+y^2-8 x=0 H : x^2 9 - y^2 4 =1 By solving x^2 9 - ( 8 x-x^2 4 )=1 aligned & 4 x^2-72 x+9 x^2=36 & 13 x^2-72 x-36=0 & 13 x^2-78 x+6 x-36=0 & 13 x(x-6)+6(x-6)=0 & x=6 or - 13 6 neglected & y^2=8(6)-(6)^2 & y= 12 aligned So, points A and B are (6, 12 ),(6,- 12 )P (h, 2 h+4 3 ) Centroid of P A B is ( 12+h 3 , 2 h+4 9 ) By options this centroid lies on the live 6 x-9 y=20