Equation of Directrix of a Parabola > Find the equation of the directrix of the parabola $y^2+4y+4x+2=0$ I tried it as follows: $$(y+2)^2+4x-2=0$$ $$(y+2)^2=-4(x-\frac{1}{2})$$ On comparing with $Y^2=-4aX, a>0$, I got $a=1$ Thus, the equation of the directrix is $X=a\Rightarrow X-1=0$. Since the origin has been shifted, the equation is $$x-\frac12-1=0$$ $$2x-3$$ However this answer is apparently wrong. This lead to another doubt. For this question, I assumed that the Directrix was parallel to the $Y$ axis. Was I correct in making this assumption? When can we/can't we make this assumption of the Directrix being parallel to one of the axes? Many thanks in anticipation.

Your work is correct. If you want you can memorize some formulas:

a parabola of equation $x=ay^2+by+c\;$ has the symmetry axis $s$ parallel to the $x$ axis and its equation is:

$s \quad: \quad y=\dfrac{-b}{2a}=h$

The vertex is at: $ \quad V\quad : \quad \left(x(h),h \right)=(k,h)$

The focus is at : $\quad F \quad :\quad (k+p,h)\; $ with $p=\dfrac{1}{4a}$

and the directrix has equation:

$d \quad : \quad x=k-p$

We can easily see that for your parabola $x=-\frac{1}{4}y^2-y-\frac{1}{2}$ the directrix is the line $x=\frac{3}{2}$.

Note that , as for all the conics , the axis of symmetry is parallel to one of the coordinate axis iff the equation does not contain a mixed term in $xy$.