Maths Olympiad Prep

Track / Stage 3 / 55 of 260 #55 of 1964

Problem 55

AMC 10/12, early questions
Geometry Difficulty 3.2 Find the answer

The distance between the focus and the directrix of the parabola y=14x2y= \frac {1}{4}x^{2} is \_\_\_\_\_.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

To solve, convert the parabola y=14x2y= \frac {1}{4}x^{2} into the standard equation form: x2=4yx^2=4y.
Therefore, the parabola opens upwards, satisfying 2p=42p=4.
Since p2=1\frac {p}{2}=1, the focus is at (0, p2\frac {p}{2}).
Thus, the coordinates of the parabola's focus are (0, 1).
Furthermore, the equation of the directrix of the parabola is y=p2y=-\frac {p}{2}, that is, y=1y=-1.
Therefore, the distance dd between the focus and the directrix of the parabola is d=1(1)=2d=1-(-1)=2.
Hence, the answer is: 2\boxed{2}.
First, convert y=14x2y= \frac {1}{4}x^{2} into the standard equation of an upward-opening parabola, obtaining the coefficient 2p=42p=4. Then, using the formula, we find the focus coordinates to be (0, 1), and the directrix equation to be y=1y=-1. Finally, we can determine the distance from the focus to the directrix of the parabola.
This question uses the parabola of a quadratic function graph as an example, focusing on the basic concepts of the parabola's focus and directrix, and is considered a basic question.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.