Question

Find the focus and the directrix for the parabola with the given equation.

Answer 1

Answer:

$\mathrm{Focus:\;\;} \left(-\frac{5}{3},\:2\right)$

$\mathrm{Directrix:\;\;}x=-\frac{1}{3}$

Step-by-step explanation:

The standard equation for a parabola is

$4p\left(x-h\right)=\left(y-k\right)^2$

with vertex at (h, k) and a focal length of |p|

$\mathrm{Rewrite}\:x=-\frac{3}{8}\left(y-2\right)^2-1\:\mathrm{in\:the\:standard\:form}$:

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}$

$x+1=-\frac{3}{8}\left(y-2\right)^2-1+1$ or

$x+1=-\frac{3}{8}\left(y-2\right)^2$

$\mathrm{Divide\:both\:sides\:by\:}-\frac{3}{8}$
$\frac{x+1}{-\frac{3}{8}}=\frac{-\frac{3}{8}\left(y-2\right)^2}{-\frac{3}{8}}$

$\mathrm{Simplify}$
$-\frac{8x}{3}-\frac{8}{3}=\left(y-2\right)^2$

$\mathrm{Factor\:}-\frac{8}{3}$
$\left(-\frac{8}{3}\right)\left(x+\frac{-\frac{8}{3}}{-\frac{8}{3}}\right)=\left(y-2\right)^2$

$\mathrm{Simplify}$
$\left(-\frac{8}{3}\right)\left(x+1\right)=\left(y-2\right)^2$

$\mathrm{Factor\:}4$
$4\cdot \frac{-\frac{8}{3}}{4}\left(x+1\right)=\left(y-2\right)^2$

$\mathrm{Simplify}$
$4\left(-\frac{2}{3}\right)\left(x+1\right)=\left(y-2\right)^2$

$\mathrm{We\; can\; rewrite\:this\;as}$
$4\left(-\frac{2}{3}\right)\left(x-\left(-1\right)\right)=\left(y-2\right)^2$

Comparing this with the standard form we get
$\left(h,\:k\right)=\left(-1,\:2\right)$
$\:p=-\frac{2}{3}$

The parabola is symmetric around the x-axis.

The focus lies a distance $p$ from the center $\left(-1,\:2\right)$ along the x axis

So focus is at
$\left(-1+p,\:2\right)$ $=\left(-1+\left(-\frac{2}{3}\right),\:2\right)$ $=\bold{ \left(-\frac{5}{3},\:2\right)}$

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

$x=-1-p$

$x=-1-\left(-\frac{2}{3}\right) = \bold{-\frac{1}{3}}$