1969 AMC 12 Problem 26

Attempt Problem 26 of the 1969 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1969 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

26.

A parabolic arch has a height of 1616 inches and a span of 4040 inches. The height, in inches, of the arch at a point 55 inches from the center M,M, is:
11
1515
151315\dfrac13
151215\dfrac12
153415\dfrac34
Answer: B
Concepts:parabolacoordinate geometrysubstitution
Small hint:
Place the center at the origin and write the parabola as y=16−ax2y=16-ax^2
Big hint:
Use the endpoint (20,0)(20,0) to find aa, then substitute x=5x=5
Solution:
Place MM at the origin with the span on the xx-axis. The arch has equation y=16−ax2.y=16-ax^2. Since (20,0)(20,0) lies on it, 0=16−400a,0=16-400a, so a=125.a=\frac{1}{25}. At x=5,x=5, y=16−2525=15. y=16-\frac{25}{25}=15.
Therefore, the correct answer is B.
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