Let ABCD be a square with the side AB lying on the line $y=x+8$. Suppose C, D lie on the parabola $x^2=y$. Find the possible values of the length of the side of the square.
I'm not sure how to start, I thought of taking the four vertices as $(t_1,t_1+8), (t_2,t_2+8), (t_3,t_3^2), (t_4,t_4^2)$ but I don't think that helps.
Let $y=x+b$ be the parallel line on which the other two vertexes lie and substitute it into $y=x^2$ to get $x^2-x-b=0$. Then, we have $x_1+x_2=1$ and $x_1x_2=-b$ and the side length $a$ of the square
$$a^2 = (y_1-y_2)^2+(x_1-x_2)^2= (x_1-x_2)^2((x_1+x_2)^2+1)\\ = ((x_1+x_2)^2-4x_1x_2)((x_1+x_2)^2+1)=2+8b $$
Note that the distance between the two parallel lines is $\frac{|8-b|}{\sqrt2}$ to establish the equation for the side length as
$$a= \sqrt{2+8b}=\frac{|8-b|}{\sqrt2}$$
Solve to obtain $b=2,\>30$ and the corresponding side lengths of the square $3\sqrt2,\> 16\sqrt2$.