The equation of the directrix of the parabola | vedclass

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(D) Given the equation of the parabola: $y^2+4y+4x+2=0$. Rearranging the terms to complete the square for $y$: $y^2+4y = -4x-2$. Adding $4$ to both si

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$x=\frac{3}{2}$

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(D) Given the equation of the parabola: $y^2+4y+4x+2=0$. Rearranging the terms to complete the square for $y$: $y^2+4y = -4x-2$. Adding $4$ to both sides: $y^2+4y+4 = -4x-2+4$. $(y+2)^2 = -4x+2$. $(y+2)^2 = -4(x-\frac{1}{2})$. Comparing this with the standard form $(y-k)^2 = -4a(x-h)$,we get: $h = \frac{1}{2}$,$k = -2$,and $4a = 4 \implies a = 1$. The equation of the directrix for a left-opening parabola is $x = h+a$. Substituting the values: $x = \frac{1}{2} + 1 = \frac{3}{2}$. Thus,the correct option is $D$.

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

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