Problem:
A segment through the focus F of a parabola with vertex V is perpendicular to FV and intersects the parabola in points A and B. What is cos(∠AVB)?
Answer Choices:
A. −735
B. −525
C. −54
D. −53
E. −21 Solution:
Let ℓ be the directrix of the parabola, and let C and D be the projections of F and B onto ℓ, respectively. For any point in the parabola, its distance to F and to ℓ are the same. Because V and B are on the parabola, it follows that p=FV=VC and 2p=FC=BD=FB. By the Pythagorean Theorem, VB=FV2+FB2=5p, and thus cos(∠FVB)=VBFV=5pp=55. Because ∠AVB=2(∠FVB), it follows that
cos(∠AVB)=2cos2(∠FVB)−1=2(55)2−1=52−1=−53
OR
Establish as in the first solution that FV=p,FB=2p, and VB=5p. Then AB=2⋅FB=4p, and by the Law of Cosines applied to △ABV,
