2015 AMC 12A Problems/Problem 12

Problem

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 $a+b$ ?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3$

Solutions

Solution 1 =

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 $a + b = \boxed{\textbf{(B)}\ 1.5}$ .

Solution 2

The parabolas must intersect the x-axis at the same two points for the kite to form. We find the x values at which they intersect by equating them and solving for x as shown below. $y = ax^2-2$ and $y = 4 - bx^2\rightarrow ax^2-2 = 4-bx^2\rightarrow (a+b)x^2 = 6 \rightarrow x = +\sqrt{\dfrac{6}{a+b}}$ or $-\sqrt{\dfrac{6}{a+b}}$ . The x-values of the y-intercepts is 0, so we plug in zero in each of them and get $-2$ and $4$ . The area of a kite is $\dfrac{d_1*d_2}{2}$ . The $d_1$ is $2+4 = 6$ . The $d_2$ is $2\sqrt{\dfrac{6}{a+b}}$ . Solving for the area $\rightarrow \dfrac{1}{2}*(6)*(2*\sqrt{\dfrac{6}{a+b}}) = 12 \rightarrow (2*\sqrt{\dfrac{6}{a+b}}) = 4 \rightarrow (\sqrt{\dfrac{6}{a+b}}) = 2 \rightarrow \dfrac{6}{a+b} = 4 \rightarrow \dfrac{6}{4} = (a+b)$ , therefore $a + b = \boxed{\textbf{(B)}\ 1.5}$ .

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2015 AMC 12A Problems/Problem 12

Contents

Problem

Solutions

Solution 1 =

Solution 2

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