Help completing square for conic

hellawowser

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Sep 25, 2017
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x^2-2x-4y^2-24y=39
In my calculations it is ((x-1)^2)/ 4 - (y-3)^2 = 1 . But using calculators and geometry programs it is said that its (y+3)^2 instead of (y-3)^2.
Please help i did it multiple times and cant understand how that 3 can be positive.
 
x^2-2x-4y^2-24y=39
In my calculations it is ((x-1)^2)/ 4 - (y-3)^2 = 1 . But using calculators and geometry programs it is said that its (y+3)^2 instead of (y-3)^2.
Please help i did it multiple times and cant understand how that 3 can be positive.
\(\displaystyle x^2 - 2x - 4y^2 - 24y = 39 \implies\)

\(\displaystyle x^2 - 2x = 4y^2 + 24y + 39 = 4(y^2 + 6y) + 39 \implies\)

\(\displaystyle x^2 - 2(1)x + 1^2 - 1^2 = 4(y^2 +2(3)y + 3^2 - 3^2) + 39 \implies \)

\(\displaystyle (x^2 - 2x + 1) - 1 = 4(y^2 + 6y + 9) - 36 + 39 \implies \)

\(\displaystyle (x - 1)^2 = 4(y + 3)^2 + 4 \implies \dfrac{(x - 1)^2}{4} = (y + 3)^2 + 1 \implies\)

\(\displaystyle \dfrac{(x - 1)^2}{4} - (y + 3)^2 = 1.\)
 
Last edited:
\(\displaystyle x^2 - 2x - 4y^2 - 24y = 39 \implies\)

\(\displaystyle x^2 - 2x = 4y^2 + 24y + 39 = 4(y^2 + 6y) + 39 \implies\)

\(\displaystyle x^2 - 2(1)x + 1^2 - 1^2 = 4(y^2 +2(3)y + 3^2 - 3^2) + 39 \implies \)

\(\displaystyle (x^2 - 2x + 1) - 1 = 4(y^2 + 6y + 9) - 36 + 39 \implies \)

\(\displaystyle (x - 1)^2 = 4(y + 3)^2 + 4 \implies \dfrac{(x - 1)^2}{4} = (y + 3)^2 + 1 \implies\)

\(\displaystyle \dfrac{(x - 1)^2}{4} - (y + 3)^2 = 1.\)

thanks
 
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