jmc

Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix. To make the algebra a bit easier, we can find the directrix of the parabola $y=8x^2,$ and then shift it upward 2 units to find the directrix of the parabola $y=8x^2+2.$

Since the parabola $y=8x^2$ is symmetric about the $y$-axis, the focus is at a point of the form $(0,f).$ Let $y=d$ be the equation of the directrix.



Let $(x,8x^2)$ be a point on the parabola $y=8x^2.$ Then $$P{F}^2=x^2+(8x^2-f{)}^2$$and $P{Q}^2=(8x^2-d{)}^2.$ Thus, $$x^2+(8x^2-f{)}^2=(8x^2-d{)}^2.$$Expanding, we get $$x^2+64x^4-16f{x}^2+f^2=64x^4-16d{x}^2+d^2.$$Matching coefficients, we get $$\begin{array}{rl}1-16f& =-16d,\\ f^2& =d^2.\end{array}$$From the first equation, $f-d=\frac{1}{16}.$ Since $f^2=d^2,$ $f=d$ or $f=-d.$ We cannot have $f=d,$ so $f=-d.$ Then $-2d=\frac{1}{16},$ so $d=-\frac{1}{32}.$

Thus, the equation of the directrix of $y=8x^2$ is $y=-\frac{1}{32},$ so the equation of the directrix of $y=8x^2+2$ is $\boxed{y=\frac{63}{32}}.$