A tangent is drawn to the parabola y2=4x at the point P whose abscissa lies in the interval [1,4] . The maximum possible area of the triangle formed by the tangent at P , the ordinates of the point P, and the x -axis is equal to
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a
8
b
16
c
24
d
32
answer is B.
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Detailed Solution
Tangent at point P is ty=x+t2, where the slope of tangent is tanθ=1/t Now, the required area is A=12(AN)(PN)=122t2(2t)=2t3=2t23/2 Now, t2∈[1,4] . Then Amax occurs when t2=4 . Therefore, Amax=16