How do you write #y^2+4x+8y+12=0# in standard form and then graph the parabola?

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Answer 1 · 2017-10-22T04:37:18

Standard form of parabola = #(y+ 4)^2 = 4* (-1) (x-1)#

graph{-(1/4)(x^2+8x+12) [-10, 10, -5, 5]}

Axis of symmetry drawn parallel to Y-axis. May rotate it by 90^0 to make it parallel to X-axis.

Standard form of parabola is #(y - k)^2 = 4p (x - h)#

#y^2 +8 y + 16 = -4x +4#

#(y+4)^2 = 4*(-1)(x - 1)#

#h = 1, k = -4, p = -1#

Focus = ((h + p), k) = (0, -4) and Directrix x = h - p = 1 - (-1) = 2#

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