Consider one sides AB of a square ABCD in order on line `y=2x-17`, and other two vertices C, D on `y=x^(2)`
The maximum possible area of square ABCD is

Options

A. 1180
B. 1250
C. 1280
D. none

Answer

C

Explanation

(3)
Let the equation of line CD be
y=2x+b (1)
and the length of side of square be a
Let the coordinates of C and D be `(x_(1),y_(1))and(x_(2),y_(2))`, respectively.
Therefore,
`(y_(2)-y_(1))/(x_(2)-x_(1))=2" "[becauseCD||AB]`
Also, `a^(2)=(y_(2)-y_(1))^(2)+(x_(2)-x_(1))^(2)=5(x_(2)-x_(1))^(2)`
`ora^(2)=5[x_(1)+(x_(2))^(2)-4x_(1)x_(2)]` (2)
Solving (1) with `y=x^(2)`, we get
`x^(2)=2x+b`
`orx_(1)+x_(2)=2,x_(1)x_(2)=-b`
Putting in (2), we get
`a^(2)=20(b+1)` (3)
Now, distance CD is

`a=|(2x_(1)-y_(1)-17)/(sqrt(5))|`
`=(|-b-17|)/(sqrt(5))` [Using (1)]
`or5a^(2)=(b+17)^(2)` (4)
From (3) and (4), we get
`100(b+1)=(b+17)^(2)`
i.e., `b=3or63`
i.e., `a^(2)=80ora^(2)=1280`