Question

How do you find the vertex, focus and directrix of `y^2+2y+12x+25=0`?

Answer 1

`(-2,-1),(-5,-1),x=1`

Explanation:

`"this is a parabola which can be expressed in the form"`

`•color(white)(x)(y-k)^2=4p(x-h)larrcolor(blue)" opens horizontally"`

`"with vertex "=(h,k)`

`• " if "4p>0" opens to the right"`

`• " if "4p<0" opens to the left"`

`color(blue)"complete the square "" on "y^2+2y`
`"moving all other terms to the right side"`

`y^2+2ycolor(red)(+1)=-12x-25color(red)(+1)`

`rArr(y+1)^2=-12(x+2)`

`rArrcolor(magenta)"vertex "=(-2,-1)`

`4p=-12rArr" opens to the left"`

`rArrp=-3`

`"p is the distance from the vertex to the focus and directrix"`

`rArr" focus "=(-2-3,-1)=(-5,-1)`

`"directrix is "x=-2+3=1`
graph{(y^2+2y+12x+25)(y-1000x+1000)((x+2)^2+(y+1)^2-0.04)((x+5)^2+(y+1)^2-0.04)=0 [-10, 10, -5, 5]}