JEE MainMathematicsParabola
A triangle has one vertex at (0, a) , where a > 0 . The other two vertices lie on the parabola y = x^2 , such that the line segment joining them is parallel to the x -axis. If the maximum possible area of this triangle is 16 , then the value of a is :
Options
- A12
- B8
- C16
- D24
Correct answer
A. 12
Step-by-step solution
Let the vertices on the parabola be (x, x^2) and (-x, x^2) for some x > 0 . The base of the triangle is 2x . The height of the triangle is a - x^2 . The area of the triangle is A(x) = 1 2 (2x)(a - x^2) = ax - x^3 . To find the maximum area, we differentiate A(x) with respect to x : A'(x) = a - 3x^2 Setting A'(x) = 0 , we get x = a 3 . The maximum area is: A_ max = a ( a 3 ) - ( a 3 )^3 = 2a 3 a 3 Given that the maximum area is 16 : 2a 3 a 3 = 16 a a = 24 3 Squaring both sides: a^3 = 576 3 = 1728 a = 12 Answer: 12