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Question 696337: Write the following in standard form and name the conic section.
1. x^2+6x-3y+15=0
2. x^2+y^2-2x+2y-7=0
and
3. x^2-6x+9+y^2-18y+81=36
I know you have to set all equations equal to zero, and complete the square i just dont know how and what order. PLEASE HELP!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Write the following in standard form and name the conic section.
1. x^2+6x-3y+15=0
x^2+6x +9 = 3y-15 + 9
(x+3)^2 = 3y-6
3(y-2) = (x+3)^2
y = (1/3)(x^3)^2+2
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Parabola opening upward from (-3,2)
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2. x^2+y^2-2x+2y-7=0
(x^2-2x+1) + (y^2+2y+1) = 7 + 1 + 1
(x-1)^2 + (y+1)^2 = 9
Circle with center at (1,-1) and radius = 3
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and
3. x^2-6x+9+y^2-18y+81=36
x^2-6x+9 + y^2-18y+81 = 36
(x-3)^2 + (y-9)^2 = 36
Circle with center at (3,9) and radius = 6
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Cheers,
Stan H.
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