Stitz-Zeager_College_Algebra_e-book

A b c x2 y 52 1 3 12 center 0 5 major axis

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 − 2y + 1 = 0. If we’re given such an equation, we can complete the square in each of the variables to see if it fits the form given in Equation 7.1 by following the steps given below. To Write the Equation of a Circle in Standard Form 1. Group the same variables together on one side of the equation and position the constant on the other side. 2. Complete the square on both variables as needed. 3. Divide both sides by the coefficient of the squares. (For circles, they will be the same.) Example 7.2.3. Complete the square to find the center and radius of 3x2 − 6x + 3y 2 + 4y − 4 = 0. Solution. 3x2 − 6x + 3y 2 + 4y − 4 3x2 − 6x + 3y 2 + 4y 4 3 x2 − 2x + 3 y 2 + y 3 4 4 3 x2 − 2x + 1 + 3 y 2 + y + 3 9 2 3 2 3 =4 factor out leading coefficients = 4 + 3(1) + 3 2 (x − 1)2 + y + add 4 to both sides 2 3(x − 1)2 + 3 y + =0 =4 4 9 complete the square in x, y = 25 3 factor = 25 9 divide both sides by 3 From Equation 7.1, we identify x − 1 as x − h, so h = 1, and y + 2 as y − k , so k = − 2 . Hence, the 3 3 2 5 center is (h, k ) = 1, − 3 . Furthermore, we see that r2 = 25 so the radius is r = 3 . 9 It is possible to o...
View Full Document

This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

Ask a homework question - tutors are online