## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.3 - The Parabola - Exercise Set - Page 997: 27

#### Answer

$x=\dfrac{1}{8}(y-2)^2+1$

#### Work Step by Step

General form of a horizontal parabola is given as: $(y-k)^2=4p(x-h)$...(1) Here, $\text{Vertex}=(h,k)$ and focus is: $(h+p, k)$ General form of a vertical parabola is given as: $(x-h)^2=4p(y-k)$...(2) Here, $\text{Vertex}=(h,k)$ and focus is: $(h, k+p)$ As we are given focus: $(3,2)$ Thus, from both forms of a parabola , we have two equations$h-p=-1$ and $h+p=3$ Add these two equations, we get: $h=1$ Thus, $p=3-h=3-1=2$ From equation (1), we have $(y-2)^2=4(2)(x-1)$ or, $(y-2)^2=8(x-1)$ or, $x-1=\dfrac{(y-2)^2}{8}$ or, $x=\dfrac{1}{8}(y-2)^2+1$

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.