Chapter 10 - Analytic Geometry - 10.1 Parabolas - 10.1 Exercises - Page 967: 38

$${x^2} = - \frac{4}{3}y$$

$$\eqalign{ & {\text{directrix }}y = \frac{1}{3} \cr & {\text{The directrix of a parabola of the form }}{x^2} = 4py{\text{ is }}y = - p \cr & \underbrace {y = \frac{1}{3}}_{y = - p},\,\,p = - \frac{1}{3} \cr & {\text{Then, the equation is:}} \cr & {x^2} = 4\left( { - \frac{1}{3}} \right)y \cr & {x^2} = - \frac{4}{3}y \cr} $$