2011 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2011 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.

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Concepts:parabola trigonometric identity Pythagorean Theorem

Difficulty rating: 1710

14.

A segment through the focus $F$ of a parabola with vertex $V$ is perpendicular to $\overline{FV}$ and intersects the parabola in points $A$ and $B.$ What is $\cos(\angle AVB)?$

$-\dfrac{3\sqrt{5}}{7}$

$-\dfrac{2\sqrt{5}}{5}$

$-\dfrac{4}{5}$

$-\dfrac{3}{5}$

$-\dfrac{1}{2}$

Solution:

Let $p=FV$ and let the directrix be $\ell.$ Projecting $F$ and $B$ onto $\ell,$ the focus-directrix property gives $FB=2p$ (the distance from $B$ to $\ell$ ), and by the Pythagorean Theorem $VB=\sqrt{FV^2+FB^2}=\sqrt{p^2+4p^2}=\sqrt5\,p.$

Then $\cos(\angle FVB)=\dfrac{FV}{VB}=\dfrac{p}{\sqrt5\,p}=\dfrac{1}{\sqrt5}.$ Since $\angle AVB=2\angle FVB,$ $\cos(\angle AVB)=2\cos^2(\angle FVB)-1=2\cdot\dfrac15-1=-\dfrac35.$

Thus, the correct answer is D.

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