2011 AMC 12B Problems/Problem 14

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 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 $\cos(\angle AVB) = -\frac{3}{5}$ $\boxed{\textbf{(D)}}$ .

(solution by mihirb)

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2011 AMC 12B Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See also