Question

What is the focus and vertex of the parabola described by `x^2+4x+4y+16=0`?

Answer 1

`"focus "=(-2,-4)," vertex "=(-2,-3)`

Explanation:

`"the equation of a vertically opening parabola is"`

`•color(white)(x)(x-h)^2=4a(y-k)`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is the distance from the vertex to the focus/directrix"`

`• " if "4a>0" then opens upwards"`

`• " if "4a<0" then opens downwards"`

`"rearrange "x^2+4x+4y+16=0" into this form"`

`"using the method of "color(blue)"completing the square"`

`x^2+4xcolor(red)(+4)=-4y-16color(red)(+4)`

`(x+2)^2=-4(y+3)`

`color(magenta)"vertex "=(-2,-3)`

`4a=-4rArra=-1`

`color(purple)" focus "=(-2,-3-1)=(-2,-4)`
graph{x^2+4x+4y+16=0 [-10, 10, -5, 5]}