Maths-
General
Easy
Question
If P (t2, 2t) t 
[0, 2] is an arbitrary point on parabola y2 = 4x. Q is foot of perpendicular from focus S on the tangent at P, then maximum area of 
PQS is-
- 1
- 2
- 5
The correct answer is: 5
Equation of tangent at P is ty = x + t2
it intersects the line x = 0 at Q
coordinates of Q are (0,t)
area of ΔPQS = 
= 
[ – t (1 – t2) + 2t] = (t + t3)
= (3t2 + 1) > 0 
t
area is maximum for t = 2
Max. area = [2 + 8] = 5
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