Answer
(a) Focus: $F(-\frac{5}{12},0)$
Directrix: $x=\frac{5}{12}$
$Focal~diameter=\frac{5}{3}$
(b)
Work Step by Step
Equation of a parabola with horizontal axis: $y^2=4px$
$5x+3y^2=0$
$3y^2=-5x$
$y^2=-\frac{5}{3}x$
$4p=-\frac{5}{3}$
$p=-\frac{5}{12}$
Focus: $F(p,0)=F(-\frac{5}{12},0)$
Directrix: $x=-p=\frac{5}{12}$
$Focal~diameter=|4p|=\frac{5}{3}$
