JEE MainMathematicsParabola
Let S be the focus and D be the directrix of the parabola x^2 = 12y . A point A lies on the parabola such that the tangent to the parabola at A makes an angle of 3 with the positive x-axis. A point B lies on the directrix D such that SA AB . If the area of SAB is K 3 , then the value of K is
Options
- A48
- B96
- C16
- D12
Correct answer
A. 48
Step-by-step solution
The equation of the parabola is x^2 = 12y . Differentiating with respect to x , we get 2x = 12 dy dx dy dx = x 6 . The tangent at A makes an angle of 3 with the positive x-axis, so its slope is ( 3 ) = 3 . Equating the slopes: x 6 = 3 x = 6 3 . Substituting x into the parabola's equation: y = (6 3 )^2 12 = 108 12 = 9 . Thus, A is (6 3 , 9) . For x^2 = 12y , 4a = 12 a = 3 . The focus is S(0, 3) and the directrix D is y = -3 . The slope of SA is m_ SA = 9 - 3 6 3 - 0 = 6 6 3 = 1 3 . Since SA AB , the slope of AB is m