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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`is1 (b) 2(c) `5/(16)`(d) 5A. 1B. 2C. `5//16`D. 5 |
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Answer» Correct Answer - D (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 |
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