JEE MainMathematicsParabola
Let P be a point in the first quadrant on the parabola y^2 = 12x , and let S be its focus. A circle drawn with the focal radius PS as diameter touches the y -axis at point M . The normal to the parabola at P intersects the x -axis at point N . If the area of the triangle SMN is 45 , then the length of PS is equal to :
Options
- A12
- B15
- C18
- D9
Correct answer
B. 15
Step-by-step solution
For the parabola y^2 = 12x , we have a = 3 . The focus is S(3, 0) . Let the coordinates of P be (3t^2, 6t) . Since P is in the first quadrant, t > 0 . The equation of the circle with PS as diameter is: (x - 3)(x - 3t^2) + (y - 0)(y - 6t) = 0 To find the point of contact M on the y -axis, put x = 0 : 9t^2 + y^2 - 6ty = 0 (y - 3t)^2 = 0 y = 3t Thus, the coordinates of M are (0, 3t) . The equation of the normal to the parabola at P(3t^2, 6t) is: y = -tx + 2at + at^3 y = -tx + 6t + 3t^3 To find the intersection N with