If `P(t^2,26),t in [0,2]` , is an arbitrary point on the parabola `y^2=4x ,Q` is the foot of perpendicular from focus `S` on the tangent at `P ,` then the maximum area of ` P Q S` is 1 (b) 2 (c) `5/(16)` (d) 5

Options

A. 1
B. 2
C. `5//16`
D. 5

Answer

D

Explanation

(4) The equation of tangent at `P(r^(2),2t)" is "ty=x+t^(2)`
It intersects the line x=0 at Q(0,t). Therefore,
`"Area of "DeltaPQS=(1)/(2)|(0,t,1),(1,0,1),(t^(2),2t,1)|`
`=(1)/(2)(t+t^(3))`
Now, `(dA)/(dt)=(1)/(2)(1+3t^(2))gt0AAtinR`
Therefore, the area is maximum for t=2. Hence, Maximum area `=(1)/(2)(2+8)=5` sq. units