SOLUTION: What is x^2+2y^2+6x-20y+53=0 in standard form?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is x^2+2y^2+6x-20y+53=0 in standard form?      Log On


   



Question 608368: What is x^2+2y^2+6x-20y+53=0 in standard form?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
x² + 2y² + 6x - 20y + 53 = 0

Rearrange the terms as

x² + 6x + 2y² - 20y = -53 

Complete the square of the first two terms:
Multiply 6 by 1%2F2 getting 3
Square 3, getting +9
Add +9 to both sides

x² + 6x + 9 + 2y² - 20y = -53+9

Factor the first three terms x²+6x+9 as (x+3)(x+3) or (x+3)²
and combine the terms on the right:

(x + 3)² + 2y² - 20y = -44 

Factor 2 out of the last two terms on the left:

(x + 3)² + 2y² - 20y = -44

(x + 3)² + 2(y² - 10y) = -44

Complete the square of the two terms inside the
second parentheses:
Multiply -10 by 1%2F2 getting -5
Square -5, getting +25
Add +25 inside the second parentheses and since there
is a 2 coefficient, we are really adding 2 times +25 or
+50 to both sides, so we add +50 to the right sides:

(x + 3)² + 2(y² - 10y + 25) = -44 + 50

Factor the three terms y²-10y+25 as (y-5)(y-5) or (y-5)²
and combine the terms on the right:

(x + 3)² + 2(y - 5)² = 6

Get 1 on the right by dividing all terms by 6

%28x+%2B+3%29%5E2%2F6 + 2%28y+-+5%29%5E2%2F6 = 6%2F6

%28x+%2B+3%29%5E2%2F6 + %28y+-+5%29%5E2%2F3 = 1

Edwin