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Difference between revisions of "2011 AMC 12B Problems/Problem 14"

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(took me a while to realize that you didn't need to memorize a bunch of equations to solve this)
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== Solution ==
 
== Solution ==
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{{solution}}
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Name the directrix of the parabola <math>l</math>. Define <math>d(X,k)</math> to be the distance between a point <math>X</math> and a line <math>k</math>.
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Now we remember the geometric definition of a parabola: given any line <math>l</math> (called the directrix) and any point <math>F</math> (called the focus), the ''parabola'' corresponding to the given directrix and focus is the locus of the points that are equidistant from <math>F</math> and <math>l</math>. Therefore <math>FV=d(V,l)</math>. Let this distance be <math>d</math>. Now note that <math>d(F,l)=2d</math>, so <math>d(A,l)=d(B,l)=2d</math>. Therefore <math>AF=BF=2d</math>. We now use the [[Pythagorean Theorem]] on triangle <math>AFV</math>; <math>AV=\sqrt{AF^2+FV^2}=d\sqrt{5}</math>. Similarly, <math>BV=d\sqrt{5}</math>. We now use the [[Law of Cosines]]:
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<cmath>AB^2=AV^2+VB^2-2AV\cdot VB\cos{\angle AVB}\Rightarrow 16d^2=10d^2-10d^2\cos{\angle AVB}</cmath>
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<cmath>\Rightarrow \cos{\angle AVB}=-\frac{3}{5}</cmath>
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This shows that the answer is <math>\boxed{\textbf{(D)}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2011|ab=B|num-b=13|num-a=15}}

Revision as of 11:09, 1 September 2011

Problem

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 $\cos\left(\angle AVB\right)$?

$\textbf{(A)}\ -\frac{3\sqrt{5}}{7} \qquad \textbf{(B)}\ -\frac{2\sqrt{5}}{5} \qquad \textbf{(C)}\ -\frac{4}{5} \qquad \textbf{(D)}\ -\frac{3}{5} \qquad \textbf{(E)}\ -\frac{1}{2}$

Solution

Name the directrix of the parabola $l$. Define $d(X,k)$ to be the distance between a point $X$ and a line $k$.

Now we remember the geometric definition of a parabola: given any line $l$ (called the directrix) and any point $F$ (called the focus), the parabola corresponding to the given directrix and focus is the locus of the points that are equidistant from $F$ and $l$. Therefore $FV=d(V,l)$. Let this distance be $d$. Now note that $d(F,l)=2d$, so $d(A,l)=d(B,l)=2d$. Therefore $AF=BF=2d$. We now use the Pythagorean Theorem on triangle $AFV$; $AV=\sqrt{AF^2+FV^2}=d\sqrt{5}$. Similarly, $BV=d\sqrt{5}$. We now use the Law of Cosines:

\[AB^2=AV^2+VB^2-2AV\cdot VB\cos{\angle AVB}\Rightarrow 16d^2=10d^2-10d^2\cos{\angle AVB}\]

\[\Rightarrow \cos{\angle AVB}=-\frac{3}{5}\]

This shows that the answer is $\boxed{\textbf{(D)}}$.

See also

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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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