2011 AMC 12B Problems/Problem 14
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[hide]Problem
A segment through the focus of a parabola with vertex is perpendicular to and intersects the parabola in points and . What is ?
Solution 1
Name the directrix of the parabola . Define to be the distance between a point and a line .
Now we remember the geometric definition of a parabola: given any line (called the directrix) and any point (called the focus), the parabola corresponding to the given directrix and focus is the locus of the points that are equidistant from and . Therefore . Let this distance be . Now note that , so . Therefore . We now use the Pythagorean Theorem on triangle ; . Similarly, . We now use the Law of Cosines:
This shows that the answer is .
Solution 2
WLOG we can assume that the parabola is . Therefore and . Also and .
and by the pythagorean theorem.
Now using the law of cosines on we have:
Thus,
(solution by mihirb)
Solution 3
After assuming that the parabola is x^2,find the points A and B, which are +/- 1,2,1/4.Now treat them as vectors,take the dot product,then find the magnitudes and multiply them.A well known definition of the dot product says that the quotient of the two is the cosine of the angle between them.This will give you D.
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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All AMC 12 Problems and Solutions |
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