If the area enclosed by the parabolas `y^2=16 a xa n dx^2=16 a y ,a >0i s(1024)/3s qdot` units, find the value of `adot`

Explanation

Given two parabolas `y^2=16 a x....(1) `
and `x^2=16 a y`
`=>y=x^2/(16a)`
now put the value of yin eqn`(1)`
`=>(x^2/(16a))^2=16ax`
`=>x^4=(16a)^3x`
`=>x[x^3-(16a)^3]=0`
`=>x=0 or x^3-(16a)^3=0`
`=>x=0 or x=16a`
`A=int_0^(16a)(y_2-y_1) `
`=int_0^(16a)(sqrt(16ax))dx-int_0^(16a)(x^2/(16a))dx`
`=4sqrt(a) [x^(3/2)/(3/2)]_0^(16a)-1/(16a)[x^3/3]_0^(16a)`
`=(8sqrt(a))/3xx64asqrt(a)-(16xx16xx16xxa^3)/(48a)`
`=(512a^2-256a^2)/3`
`=((256a^2)/3)` sq units.
according to question,
`(256a^2)/3=(1024)/3`
`a=pm2`