Elementary Geometry for College Students (7th Edition) Clone

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: 46

Answer

$y = \frac{1}{8}~x^2$

Work Step by Step

We can write an expression for the distance from a point $(x,y)$ to the focus $(0,2)$: $d_1 = \sqrt{(x-0)^2+(y-2)^2}$ $d_1 = \sqrt{x^2+(y-2)^2}$ We can write an expression for the distance from a point $(x,y)$ to the directrix $y = -2$: $d_2 = \sqrt{(0)^2+[y-(-2)]^2}$ $d_2 = \sqrt{(y+2)^2}$ $d_2 = y+2$ We can equate these two distances to find the equation of the parabola: $d_2 = d_1$ $y+2 = \sqrt{x^2+(y-2)^2}$ $(y+2)^2 = x^2+(y-2)^2$ $y^2+4y+4 = x^2+y^2-4y+4$ $8y = x^2$ $y = \frac{1}{8}~x^2$
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