Question

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

Answer 1

`\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
`p` 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)}`