The equation of the directrix of the parabola | vedclass

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Question

(D) Given equation: ${y^2} + 4y + 4x + 2 = 0$Rewrite as: ${y^2} + 4y + 4 = -4x + 2$${(y + 2)^2} = -4x + 2$${(y + 2)^2} = -4(x - \frac{1}{2})$Comparing

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given equation: ${y^2} + 4y + 4x + 2 = 0$Rewrite as: ${y^2} + 4y + 4 = -4x + 2$${(y + 2)^2} = -4x + 2$${(y + 2)^2} = -4(x - \frac{1}{2})$Comparing with standard form ${Y^2} = -4aX$,where $Y = y + 2$,$X = x - \frac{1}{2}$,and $4a = 4 \implies a = 1$.The directrix of ${Y^2} = -4aX$ is $X = a$.Substituting the values: $x - \frac{1}{2} = 1$$x = 1 + \frac{1}{2} = \frac{3}{2}$.

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

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