We proceed to complete the squares in x and in y. x²2²-2x+4y² + 8y + 1 = 0 x² - 2x + 4y² + 8y = -1 (x²-2x) +4(y² + 2y) = -1 Place the constant on the right side. Factor out 4 from the y-terms.
We proceed to complete the squares in x and in y. x²2²-2x+4y² + 8y + 1 = 0 x² - 2x + 4y² + 8y = -1 (x²-2x) +4(y² + 2y) = -1 Place the constant on the right side. Factor out 4 from the y-terms.
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How is the square completed?
![We proceed to complete the squares in \( x \) and in \( y \).
\[ x^2 - 2x + 4y^2 + 8y + 1 = 0 \]
Place the constant on the right side:
\[ x^2 - 2x + 4y^2 + 8y = -1 \]
Factor out 4 from the \( y \)-terms:
\[ (x^2 - 2x) + 4(y^2 + 2y) = -1 \]
Complete each square:
\[ (x^2 - 2x + 1) + 4(y^2 + 2y + 1) = -1 + 1 + 4 \cdot 1 \]
Factor:
\[ (x-1)^2 + 4(y+1)^2 = 4 \]
Divide by 4:
\[ \frac{(x-1)^2}{4} + \frac{4(y+1)^2}{4} = \frac{4}{4} \]
Simplify:
\[ \frac{(x-1)^2}{4} + (y+1)^2 = 1 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff462e736-7a38-4abe-8728-e08b1ac68dd9%2F7133866a-07f4-4fc5-bbba-7250d7dea384%2Fwox72th_processed.jpeg&w=3840&q=75)
Transcribed Image Text:We proceed to complete the squares in \( x \) and in \( y \).
\[ x^2 - 2x + 4y^2 + 8y + 1 = 0 \]
Place the constant on the right side:
\[ x^2 - 2x + 4y^2 + 8y = -1 \]
Factor out 4 from the \( y \)-terms:
\[ (x^2 - 2x) + 4(y^2 + 2y) = -1 \]
Complete each square:
\[ (x^2 - 2x + 1) + 4(y^2 + 2y + 1) = -1 + 1 + 4 \cdot 1 \]
Factor:
\[ (x-1)^2 + 4(y+1)^2 = 4 \]
Divide by 4:
\[ \frac{(x-1)^2}{4} + \frac{4(y+1)^2}{4} = \frac{4}{4} \]
Simplify:
\[ \frac{(x-1)^2}{4} + (y+1)^2 = 1 \]
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