\[\begin{array}{1 1}(1)x=\frac{15}{4}&(2)x=-\frac{15}{4}\\(3)\frac{17}{4}&(4)x=-\frac{17}{4}\end{array}\]

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$y^2=x+4$

$(y-0)^2=(x+4)$

$y^2=X$ Where $Y= y-0 \;X=x+4$

$Y^2= 4 \bigg( \large\frac{1}{4} \bigg) $$x$

$a= \large\frac{1}{4}$

The equation of the directrix is

$x= -\large\frac{1}{4}$

$x+4=-\large\frac{1}{4}$

$x= -\large\frac{1}{4}$$-4 $

$\quad= \large\frac{-1-16}{4}$

$\quad= \large\frac{-17}{4}$

Hence 3 is the correct answer.

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