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Consider the parabola y^(2)=4x, let P an...

Consider the parabola `y^(2)=4x`, let P and Q be two points `(4,-4)` and `(9,6)` on the parabola. Let R be a moving point on the arc of the parabola whose x-coordinate is between P and Q. If the maximum area of triangle PQR is K, then `(4K)^(1//3)` is equal to

A

`(125)/(2)`

B

`(75)/(2)`

C

`(625)/(4)`

D

`(125)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given parabola is `y^(2)=4x,`
Since, X lies on the parabola, so let the fcoordinates of X be `(t^(2),2t).` Thus the coordinates of the vertices of the traingle PXQ are `P(4,4),x(t^(2)2t)` and Q (9,6).

`therefore"Area of" DeltaPXQ=1/2||{:(4,-4,1),(t^(2),2t,1),(9,6,1):}||`
`=1/2|"["4(2t-6)+4(t^(2)9-)+1(6t^(2)-18t)"]"|`
`=1/2|"["8t-24+4t^(2)-36+6t^(2)-18t"]"|`
`=|5t^(2)-5t-30|=|5(t+2)(t-3)|`
Now, as X is any point on the are POQ of the parabola, therefore ordinate of point `X, 2t in (-4,6)impliest in (-2,3).`
`therefore"Area of"DeltaPXQ=-5(t+2)(t-3)=-5t^(2)+5t+5t+30`
`" "[because|x-a|=-(x-a).ifx lt a]`
The maximum area (in square units)
`=-[(25-4(-5)(30))/(-(-5))]=(125)/(4)`
[`because` Maximum value of quadratic expression `ax^(2)+bx+c, when a lt 0 is -(D)/(4alpha)"]"`
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