BITSAT2022MathematicsParabolaActual
A normal is drawn at the point P to the parabola y^2=8 x , which is inclined at 60^ with the straight line y=8 . Then the point P lies on the straight line
Options
- A2 x+y-12-4 3 =0
- B2 x-y-12+4 3 =0
- C2 x-y-12-4 3 =0
- DNone of these
Correct answer
C. 2 x-y-12-4 3 =0
Step-by-step solution
For the parabola y^2=4 a x , the equation of normal at P (a m^2,-2 a m ) is y=m x-2 a m-a m^3 . Here, m= 60^ = 3 P (a( 3 )^2,-2 a( 3 ) ) (6,-4 3 ) ( a=2) Thus, P satisfies 2 x-y-12-4 3 =0