2021 AMC 12A Spring Problem 20

Question

Suppose that on a parabola with vertex

V

and a focus

F

there exists a point

A

such that

AF = 20

and

AV = 21.

What is the sum of all possible values of the length

FV?

13

\dfrac{40}{3}

\dfrac{41}{3}

14

\dfrac{43}{3}

Put

V

at the origin with focal distance

f = FV;

then

AF

equals the distance from

A

to the directrix

With

A = (x, y),

use

y + f = 20

and

x^2 + y^2 = 21^2,

with

x^2 = 4fy,

to get a quadratic in

f

Let

V = (0, 0),

focus

F = (0, f),

and directrix

y = -f,

where

f = FV.

A point

A = (x, y)

on the parabola satisfies

x^2 = 4fy

and

AF = y + f = 20,

so

y = 20 - f.

Also

AV^2 = x^2 + y^2 = 4fy + y^2

= 441.

Substituting

y = 20 - f:

\begin{aligned} &4f(20 - f) + (20 - f)^2 = 441 \\ &\;\Longrightarrow\; 3f^2 - 40f + 41 = 0. \end{aligned}

By Vieta’s formulas, the sum of the two possible values of

f

is

\dfrac{40}{3}.

Thus, the correct answer is B.

Problem 20 in other years