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:

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

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

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

(solution by mihirb)

Solution 3

After assuming that the parabola is x^2,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 $y-k=\frac{1}{4a}(x-h)^2$ , where $(h,k)$ is the vertex and $a$ is the distance from the focus to the vertex, here $F$ to $V$ . The length of the latus rectum, or $\overline{AB}$ here, is equal to $|4a|$ . This means that $AF=BF=2a$ since $F$ is the midpoint of $AB$ . Then we can use right Triangle $VAF$ to figure out that $VA=\frac{a\sqrt{5}}{4}$ . Now we can use the fact that $\cos{\angle AVB}=\cos{2 \angle AVF}$ and use the double angle formula. This results in $\cos{2 \angle AVF}=2{\cos}^2{\angle AVF}-1$ . We can find $\cos{\angle AVF}$ from right triangle $AVF$ using the Pythagorean Theorem, which is $\frac{a}{\sqrt{5}}$ . Evaulating the expression, we find that $\cos{2 \angle AVF}=\frac{-3}{5}$ .

-Indefintense

= 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search