Question

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 @

Answer 1

@Nurali please help

Answer 2

(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

Answer 3

can u help me with aother one? @Nurali

Answer 4

i will try.

Answer 5

Find 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

Answer 6

@Nurali

Answer 7

@Nurali please help