`L L '` is the latus sectum of the parabola `y^2=4a xa n dP P '` is a double ordinate drawn between the vertex and the latus rectum. Show that the area of the trapezium `P P^(prime)L L '` is maximum when the distance `P P '` from the vertex is `a//9.`

Explanation

Let LL' = 4a be the latus rectangular of the parabola `y^(2)` =4ax and let `at^(2)`,2at be the coordinates of the point P.

Here PP is the double ordinate of the parabola.Then
`OM=at^(2) "or" MN =ON-OM =-at^(2)`
and PP=2PM=4at
Now area of trapezium PP'LL
`A=(1)/(2)(PP'+LL')xxMN`
`=(1)/(2)(4at+4a)(a-at^(2))`
`=2a^(2)(-t^(3)-t^(2)+t+1)`
or `dA//dt=2a^(2)(-3t^(2)-2t+1)` and `A//dt^(2)=2a^(2)(-6t-2)`
For maxium or minimum of `A,dA//dt=0`
or `-2a^(12)(3t-1)(t+1)=0`
or `t=-1,1//3`
when `t=-1,d^(2)A//dt^(2)=8a^(2)(+ve)`
Thus A is minimum when t=-1
when `t=1//3,d^(2)A//dt^(2)=-8a^(2)(-ve)`
Thus A is maximium when `t=1//3` (only point of maxima)
Therefore , for the area of trapezium PP'LL to be maximum distance of PP' from vertex O is OM =`at^(2)=a(1//3)^(2)=a//9`