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 $FV?$

$13$

$\dfrac{40}{3}$

$\dfrac{41}{3}$

$14$

$\dfrac{43}{3}$

Answer: B

Concepts:parabola distance formula Vieta’s Formulas

Difficulty rating: 2300

Small Hint:

Put $V$ at the origin with focal distance $f = FV;$ then $AF$ equals the distance from $A$ to the directrix

Big Hint:

With $A = (x, y),$ use $y + f = 20$ and $x^2 + y^2 = 21^2,$ with $x^2 = 4fy,$ to get a quadratic in $f$

Solution:

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.

2021 AMC 12A Spring Problem 20

Answer: B

Difficulty rating: 2300

Small Hint:

Big Hint:

Solution:

Problem 20 in Other Years