The directrix of the parabola {x^2 - 4x - 8y | vedclass

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Question

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

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$y = -1$

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(D) The given equation of the parabola is ${x^2 - 4x - 8y + 12 = 0}$.Completing the square for the $x$ terms:${x^2 - 4x + 4 = 8y - 12 + 4}$${(x - 2)^2 = 8y - 8}$${(x - 2)^2 = 8(y - 1)}$Comparing this with the standard form ${X^2 = 4aY}$,where ${X = x - 2}$,${Y = y - 1}$,and ${4a = 8}$,we get ${a = 2}$.The equation of the directrix for the parabola ${X^2 = 4aY}$ is ${Y = -a}$.Substituting the values back:${y - 1 = -2}$${y = -1}$.

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

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