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Difference between revisions of "2015 AMC 12A Problems/Problem 12"
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==Solution==
 
==Solution==
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Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by <math>4 - (-2)</math> (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 <math>(2, 0), (-2, 0).</math> Then <math>0 = 4a - 2 \rightarrow a = 0.5</math>, and <math>0 = 4 - 4b \rightarrow b = 1.</math> Then <math>a + b = 1.5</math>, or <math>\textbf{(B)}</math>.
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Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by <math>4 - (-2)</math> (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 <math>(2, 0), (-2, 0).</math> Then <math>0 = 4a - 2 \rightarrow a = 0.5</math>, and <math>0 = 4 - 4b \rightarrow b = 1.</math> Then <math>a + b = \boxed{\textbf{(B)}\ 1.5}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2015|ab=A|num-b=11|num-a=13}}

Revision as of 01:04, 11 February 2017

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 $a+b$?

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


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

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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