The area of the trapezium whose vertices lie on the parabola `y^2 = 4x` and its diagonals pass through (1,0) and having length `25/4` units each is

Explanation

Clearly, trapezium ABCD inscribed in the parabola will be isosceles as shown in the following figure.

Let point A be `(t^(2),2t)`.
`:." "AS=1+t^(2)` (focal distance)
`:." "CS=AC-AS=(25)/(4)-(1+t^(2))=(21)/(4)-t^(2)`
Also semi latus rectum is harmonic mean of segments AS and CS
`:." "(1)/(AS)+(1)/(CS)=(1)/(a)`
`rArr" "(1)/(1+t_(2))+(1)/((21)/(4)-t^(2))=1`
`rArr" "(1)/(1+t^(2))+(4)/(21-4t^(2))=1`
`rArr" "21-4t^(2)+4+4t^(2)=21+17t^(2)-4t^(4)`
`rArr" "4t^(4)-17t^(2)+4=0`
`rArr" "(4t^(2)-1)(t^(2)-4)=0`
`rArr" "t^(2)=4,(1)/(4)`
`rArr" "t=pm2,pm(1)/(2)`
Thus, `A-=(1//4,1),B-=(4,4),C-=(4,-4)andD-=(1//4,-1)`
`:." "AD=2`
And BC = 8 and distance between AD and BC `=4-(1)/(4)=(15)/(4)`
`:.` Area of trapezium `ABCD=(1)/(2)(2+8)xx(15)/(4)=(75)/(4)` sq. units