If $x+a=0$ is the directrix of the parabola $y^2=2y+ax+2$ then the positive value of a is

$$\begin{gathered} y^2-2y+1=ax+3 \\ {(y-1)}^2=a(x+\frac{3}{a})\to \left(1\right) \\ \alpha =\frac{-3}{a},\beta =1 \end{gathered}$$

From equation (1) directrix equation is $=4a^1$

$\Rightarrow 4a^1=a\Rightarrow a^1=\frac{a}{4}$

From equation (1) directix equation is $x=\alpha -a^1$

$x=\frac{-3}{a}-\frac{a}{4}\Rightarrow x=\frac{-12-a^2}{4a}\to \left(2\right)$

Given directrix is $x+a=0\Rightarrow x=-a\to \left(3\right)$

Form (2)and(3)

$-a=\frac{-12-a^2}{4a}\Rightarrow 4a^2=12+a^2$

$3a^2=12\Rightarrow a^2=4\Rightarrow a=\pm 2$

$\therefore$positive value of a is 2