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Difference between revisions of "2011 AMC 12B Problems/Problem 14"
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<math>\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}</math>
 
<math>\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}</math>
  
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== Solution ==
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== Solution 1==
 
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>.
 
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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This shows that the answer is <math>\boxed{\textbf{(D)}}</math>.
 
This shows that the answer is <math>\boxed{\textbf{(D)}}</math>.
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==Solution of 2==
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WLOG we can assume that the parabola is <math>y=x^2</math>. Therefore <math>V = (0,0)</math> and <math>F = (0,\frac{1}{4})</math>. Also <math>A = (-\frac{1}{2},\frac{1}{4})</math> and <math>B = (\frac{1}{2},\frac{1}{4})</math>.
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<math>AB = 1</math> and <math>AV = VB = \sqrt{(\frac{1}{2})^2+(\frac{1}{4})^2} = \frac{\sqrt{5}}{4}</math> by the pythagorean theorem.
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Now using the law of cosines on <math>\triangle AVB</math> we have:
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<math>AB^2 = 2AV^2-2AV\cos(\angle AVB) = 2AV^2(1-\cos(\angle AVB))</math>
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<math>1 = \frac{5}{8} \cos(\angle AVB)</math>
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and <math>\cos(\angle AVB) = -\frac{3}{5}</math> as desired.
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:23, 21 March 2015

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 1

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 $\boxed{\textbf{(D)}}$.


Solution of 2

WLOG we can assume that the parabola is $y=x^2$. Therefore $V = (0,0)$ and $F = (0,\frac{1}{4})$. Also $A = (-\frac{1}{2},\frac{1}{4})$ and $B = (\frac{1}{2},\frac{1}{4})$.

$AB = 1$ and $AV = VB = \sqrt{(\frac{1}{2})^2+(\frac{1}{4})^2} = \frac{\sqrt{5}}{4}$ by the pythagorean theorem.

Now using the law of cosines on $\triangle AVB$ we have:

$AB^2 = 2AV^2-2AV\cos(\angle AVB) = 2AV^2(1-\cos(\angle AVB))$

$1 = \frac{5}{8} \cos(\angle AVB)$

and $\cos(\angle AVB) = -\frac{3}{5}$ as desired.

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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