The equation of the directrix of the parabola | vedclass

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(B) Given parabola is: $y^2 - x + 4y + 5 = 0$ $y^2 + 4y = x - 5$ Adding $4$ to both sides to complete the square: $y^2 + 4y + 4 = x - 5 + 4$ $(y + 2)^

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$4x - 3 = 0$

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(B) Given parabola is: $y^2 - x + 4y + 5 = 0$ $y^2 + 4y = x - 5$ Adding $4$ to both sides to complete the square: $y^2 + 4y + 4 = x - 5 + 4$ $(y + 2)^2 = (x - 1)$ Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get: Vertex $(h, k) = (1, -2)$ and $4a = 1$,so $a = \frac{1}{4}$. The equation of the directrix for a parabola of the form $(y - k)^2 = 4a(x - h)$ is $X = -a$,where $X = x - h$. Substituting the values: $x - 1 = -\frac{1}{4}$ $x = 1 - \frac{1}{4} = \frac{3}{4}$ $4x = 3$ $4x - 3 = 0$

The equation of the directrix of the parabola $y^2-x+4y+5=0$ is

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