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

Question

asked Sep 5, 2023 196k views

1 Answer

Cyupa · Answer 1 · 2023-09-09T14:15:04+0000

Answer:

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

$\mathrm{Directrix:\;\;}x=-(1)/(3)$

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|

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

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

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

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

$\mathrm{Divide\:both\:sides\:by\:}-(3)/(8)$

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

$\mathrm{Simplify}$

$-(8x)/(3)-(8)/(3)=\left(y-2\right)^2$

$\mathrm{Factor\:}-(8)/(3)$

$\left(-(8)/(3)\right)\left(x+(-(8)/(3))/(-(8)/(3))\right)=\left(y-2\right)^2$

$\mathrm{Simplify}$

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

$\mathrm{Factor\:}4$

$4\cdot (-(8)/(3))/(4)\left(x+1\right)=\left(y-2\right)^2$

$\mathrm{Simplify}$

$4\left(-(2)/(3)\right)\left(x+1\right)=\left(y-2\right)^2$

$\mathrm{We\; can\; rewrite\:this\;as}$

$4\left(-(2)/(3)\right)\left(x-\left(-1\right)\right)=\left(y-2\right)^2$

Comparing this with the standard form we get

$\left(h,\:k\right)=\left(-1,\:2\right)$

$\:p=-(2)/(3)$

The parabola is symmetric around the x-axis.

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

So focus is at

$\left(-1+p,\:2\right)$
$=\left(-1+\left(-(2)/(3)\right),\:2\right)$
$=\bold{ \left(-(5)/(3),\:2\right)}$

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

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