MATH SOLVE

8 months ago

Q:
# 7x2 + 7y2 − 28x + 42y − 35 = 0. in standard form

Accepted Solution

A:

Answer:[tex](x-2)^2+(y+3)^2=18[/tex]Step-by-step explanation:This is a circle. It already is in standard form, so maybe you mean work form? I'll go with that, since there's nowhere else to go!Let's group the x terms together and the y terms together and move the constant to the other side of the equals sign first.[tex]7x^2-28x+7y^2+42y=35[/tex]In the world of conics, this is a circle because the leading coefficients on the x-squared term and y-squared term are the same, namely 7.But if we are going to get this into its graphing form, or work form, we need that 7 to be a 1. We are going to complete the square on both the x terms and the y terms. First we factor out the 7 from both x and y:[tex]7(x^2-4x)+7(y^2+6y)=35[/tex]The rules for completing the square are as follows:Take half the linear term, square it, and add it to both sides, keeping in mind the 7 you factored out as a multiplier.Our linear term in the first set of parenthesis is 4 (from the -4x). Half of 4 is 2, and 2 squared is 4. We add a 4 into the parenthesis, but multiplying in the 7 has us actually adding in a 28.Our linear term in the second set of parenthesis is a (from 6y). Half of 6 is 3, and 3 squared is 9. We add a 9 into the parenthesis, but multiplying in the 7 has us actually adding in a 63. Now here is what we have done thus far:[tex]7(x^2-4x+4)+7(y^2+6y+9)=35+28+63[/tex]We have, in the process of completing the square, created 2 perfect square binomials on the left. That's why we do this...to get those perfect square binomials. Stating those on the left and adding on the right breaks down our equation into work form. Well, almost.[tex]7(x-2)^2+7(y+3)^2=126[/tex]Actual work form has us divide both sides by 7 to get the final answer of[tex](x-2)^2+(y+3)^2=18[/tex]And there you are!