Elementary Geometry for College Students (7th Edition)

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 451: 45

Answer

$y = \frac{1}{4}~x^2$

Work Step by Step

We can write an expression for the distance from a point $(x,y)$ to the focus $(0,1)$: $d_1 = \sqrt{(x-0)^2+(y-1)^2}$ $d_1 = \sqrt{x^2+(y-1)^2}$ We can write an expression for the distance from a point $(x,y)$ to the directrix $y = -1$: $d_2 = \sqrt{(0)^2+[y-(-1)]^2}$ $d_2 = \sqrt{(y+1)^2}$ $d_2 = y+1$ We can equate these two distances to find the equation of the parabola: $d_2 = d_1$ $y+1 = \sqrt{x^2+(y-1)^2}$ $(y+1)^2 = x^2+(y-1)^2$ $y^2+2y+1 = x^2+y^2-2y+1$ $4y = x^2$ $y = \frac{1}{4}~x^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.