The equation of the directrix for the parabol | vedclass

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Question

(A) Given the parabola equation: $4x^{2}-4x-2y+3=0$ $4(x^{2}-x) = 2y-3$ $4(x^{2}-x+\frac{1}{4}-\frac{1}{4}) = 2y-3$ $4(x-\frac{1}{2})^{2}-1 = 2y-3$ $4

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$2y=1$

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(A) Given the parabola equation: $4x^{2}-4x-2y+3=0$ $4(x^{2}-x) = 2y-3$ $4(x^{2}-x+\frac{1}{4}-\frac{1}{4}) = 2y-3$ $4(x-\frac{1}{2})^{2}-1 = 2y-3$ $4(x-\frac{1}{2})^{2} = 2y-2$ $(x-\frac{1}{2})^{2} = \frac{1}{2}(y-1)$ Comparing this with the standard form $X^{2} = 4aY$,where $X = x-\frac{1}{2}$ and $Y = y-1$: $4a = \frac{1}{2} \implies a = \frac{1}{8}$ The equation of the directrix is $Y+a = 0$ $y-1+\frac{1}{8} = 0$ $y = \frac{7}{8}$ Wait,re-evaluating the standard form: $(x-\frac{1}{2})^{2} = 2(y-1)$ implies $4a = 2$,so $a = \frac{1}{2}$. The directrix is $Y = -a \implies y-1 = -\frac{1}{2}$ $y = 1 - \frac{1}{2} = \frac{1}{2}$ $2y = 1$.

The equation of the directrix for the parabola $4x^{2}-4x-2y+3=0$ is

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