Answer:
[tex]\mathrm{Focus:\;\;} \left(-\frac{5}{3},\:2\right)[/tex]
[tex]\mathrm{Directrix:\;\;}x=-\frac{1}{3}[/tex]
Step-by-step explanation:
The standard equation for a parabola is
[tex]4p\left(x-h\right)=\left(y-k\right)^2[/tex]
with vertex at (h, k) and a focal length of |p|
[tex]\mathrm{Rewrite}\:x=-\frac{3}{8}\left(y-2\right)^2-1\:\mathrm{in\:the\:standard\:form}[/tex]:
[tex]\mathrm{Add\:}1\mathrm{\:to\:both\:sides}[/tex]
[tex]x+1=-\frac{3}{8}\left(y-2\right)^2-1+1[/tex] or
[tex]x+1=-\frac{3}{8}\left(y-2\right)^2[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}-\frac{3}{8}[/tex]
[tex]\frac{x+1}{-\frac{3}{8}}=\frac{-\frac{3}{8}\left(y-2\right)^2}{-\frac{3}{8}}[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]-\frac{8x}{3}-\frac{8}{3}=\left(y-2\right)^2[/tex]
[tex]\mathrm{Factor\:}-\frac{8}{3}[/tex]
[tex]\left(-\frac{8}{3}\right)\left(x+\frac{-\frac{8}{3}}{-\frac{8}{3}}\right)=\left(y-2\right)^2[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]\left(-\frac{8}{3}\right)\left(x+1\right)=\left(y-2\right)^2[/tex]
[tex]\mathrm{Factor\:}4[/tex]
[tex]4\cdot \frac{-\frac{8}{3}}{4}\left(x+1\right)=\left(y-2\right)^2[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]4\left(-\frac{2}{3}\right)\left(x+1\right)=\left(y-2\right)^2[/tex]
[tex]\mathrm{We\; can\; rewrite\:this\;as}[/tex]
[tex]4\left(-\frac{2}{3}\right)\left(x-\left(-1\right)\right)=\left(y-2\right)^2[/tex]
Comparing this with the standard form we get
[tex]\left(h,\:k\right)=\left(-1,\:2\right)[/tex]
[tex]\:p=-\frac{2}{3}[/tex]
The parabola is symmetric around the x-axis.
The focus lies a distance [tex]p[/tex] from the center [tex]\left(-1,\:2\right)[/tex] along the x axis
So focus is at
[tex]\left(-1+p,\:2\right)[/tex] [tex]=\left(-1+\left(-\frac{2}{3}\right),\:2\right)[/tex] [tex]=\bold{ \left(-\frac{5}{3},\:2\right)}[/tex]
The parabola is symmetric around the x-axis and so the directrix is a line parallel to the y-axis at a distance -p from the center
[tex]x=-1-p[/tex]
[tex]x=-1-\left(-\frac{2}{3}\right) = \bold{-\frac{1}{3}}[/tex]
