Suppose I have two parabolas defined by the following equations

$\displaystyle f_1 = -x^2 + bx + c$ which has vertex $\displaystyle V_1 = (\frac{b}{2} , c + \frac{b^2}{4})$

$\displaystyle f_2 = x^2$ which has vertex $\displaystyle V_2 = (0,0)$

For some values $\displaystyle b, c$, these parabolas intersect as points $\displaystyle R, S$

The first question asks

(i) Determine all pairs $\displaystyle (b,c)$ for which the points $\displaystyle R, S$ exist and the points $\displaystyle V_1, V_2, R, S$ are distinct.

Setting $\displaystyle f_1 = f_2$ results in $\displaystyle 2x^2- bx - c = 0$, and two distinct solutions x exist when the discriminant of the quadratic is greater than 0.

Thus, it's all pairs $\displaystyle (b,c)$ such that $\displaystyle D = b^2 + 8c > 0$

(ii) Determine all pairs $\displaystyle (b,c)$ for which the points R and S exist, the points $\displaystyle V_1,V_2,R,S$ are distinct, and quadrilateral $\displaystyle V_1RV_2S$ is a rectangle.

This is where I'm a bit stumped. I think I would still have a range of solutions as (i) and that I would use some of these properties of rectangles.

* Opposite sides have equal slope

* Adjacent sides have negative reciprocal slope

* The intersection of diagonals occurs at the midpoint.