AMC12 2015 12A, Problem 12

The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$. What is $a+b$?

Solution: Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by $4 - (-2)$ (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are $(2, 0), (-2, 0).$ Then $0 = 4a - 2 \rightarrow a = 0.5$, and $0 = 4 - 4b \rightarrow b = 1.$ Then $a + b = \fbox{1.5}$.

Answer: 1.5