2011 AMC 12B Problem 14

Question

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}

Let

p=FV.

Use the focus-directrix property to find

FB

and then

VB

Find

\cos(\angle FVB),

then

\angle AVB=2\angle FVB

via the double angle formula

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

\begin{aligned} VB&=\sqrt{FV^2+FB^2} \\ &=\sqrt{p^2+4p^2}=\sqrt5\,p. \end{aligned}

Then

\cos(\angle FVB)=\dfrac{FV}{VB}

=\dfrac{p}{\sqrt5\,p}

=\dfrac{1}{\sqrt5}.

Since

\angle AVB=2\angle FVB,

\begin{aligned} \cos(\angle AVB) &=2\cos^2(\angle FVB) \\ &\quad {}-1 \\ &=2\cdot\dfrac15-1 \\ &=-\dfrac35. \end{aligned}

Thus, the correct answer is D.

Problem 14 in other years