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anonymous
 one year ago
Find the standard form of the equation of the parabola with a focus at (0, 3) and a directrix at y = 3.
y2 = 12x
y2 = 3x
y = negative 1 divided by 12x2
y = negative 1 divided by 3x2
@ganeshie8 @dan815 @solomonzelman @nurali @
anonymous
 one year ago
Find the standard form of the equation of the parabola with a focus at (0, 3) and a directrix at y = 3. y2 = 12x y2 = 3x y = negative 1 divided by 12x2 y = negative 1 divided by 3x2 @ganeshie8 @dan815 @solomonzelman @nurali @

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(x  x)^2 + (y  3)^2 = (x  0)^2 + (y  3)^2 y^2 6y + 9 = x^2 + y^2 + 6y + 9 12 y = x^2 y =  x^2 / 12 (or (1/12)x^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u help me with aother one? @Nurali

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3 divided by 2.x. y squared over 36 minus x squared over 16 = 1 y squared over 4 minus x squared over 9 = 1 y squared over 36 minus x squared over 4 = 1 y squared over 16 minus x squared over 36 = 1
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