One side of square `ABCD` is on the line `y=2x-17` and other two vertices are on parabola `y=x^(2)`, if the minimum area of square is `A` then find `A/16`.

Answer

5

Explanation

`BC=(|2t-t^(2)-17|)/(sqrt(5))=a`
`impliest^(2)-2t+17-sqrt(5)a=0`……..i
If `C(t_(1),t_(1)^(2)),D(t_(2),t_(2)^(2))`, then
`t_(1),t_(2)` are roots of equation i and `CD=a`
`impliest_(1)+t_(2)=2,t_(1)t_(2)=17-sqrt(5)a`
`impliesa^(2)=(t_(1)-t_(2))^(2)+(t_(1)^(2)-t_(2)^(2))^(2)`
`a^(2)=4-4(17-sqrt(5)a)+4(4-4(17-sqrt(5)a))`
`a^(2)=20-20(17)+20sqrt(5)a`
`impliesa^(2)=20sqrt(5)a+320=0`
`impliesa=16sqrt(5)` or `a=4sqrt(5)`
Minimum area is when `a=4sqrt(5)impliesa^(2)=80`
`impliesA/16=5`