## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 443: 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: $d_2 = d_1$ $y+1 = \sqrt{x^2+(y-1)^2}$ $(y+1)^2 = x^2+(y-1)^2$ $y^2+2y+1 = x^2+y^2-2y+1$ $4y = x^2$ $y = \frac{1}{4}~x^2$

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