2011 AMC 12B Problems/Problem 14
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 ,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.
Solution 4
As we know, an expression for the equation of a parabola is , where is the vertex and is the distance from the focus to the vertex, here to . The length of the latus rectum, or here, is equal to . This means that since is the midpoint of . Then we can use right Triangle to figure out that . Now we can use the fact that and use the double angle formula. This results in . We can find from right triangle using the Pythagorean Theorem, which is . Evaulating the expression, we find that .
-Indefintense
= See also ==
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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