The equation of the directrix of the parabola | vedclass

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(C) The given equation of the parabola is $x^2+8x+12y+4=0$. Completing the square for the $x$ terms: $x^2+8x = -12y-4$ $(x+4)^2 - 16 = -12y-4$ $(x+4)^

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$y-4=0$

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(C) The given equation of the parabola is $x^2+8x+12y+4=0$. Completing the square for the $x$ terms: $x^2+8x = -12y-4$ $(x+4)^2 - 16 = -12y-4$ $(x+4)^2 = -12y+12$ $(x+4)^2 = -12(y-1)$. Comparing this with the standard form $(x-h)^2 = 4a(y-k)$,we get $4a = -12$,so $a = -3$. The vertex is $(h, k) = (-4, 1)$. The equation of the directrix for a downward-opening parabola is $y = k - a$. Substituting the values: $y = 1 - (-3) = 1 + 3 = 4$. Thus,the equation of the directrix is $y-4=0$.

The equation of the directrix of the parabola $x^2+8x+12y+4=0$ is

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