Question

What are the vertex, focus, and directrix of the parabola with the equation x^2+8x+4y+4=0

Answer 1

We are given the function x^2+8x+4y+4=0. To determine the characteristics of this function, we need to write it in the standard form as follows:

x^2+8x+4y+4=0
4y = -x^2 - 8x - 4
y = (-1/4)x^2 - 2x - 1

To determine the vertex and the focus of the parabola, we write it in the form (y+k)^2 = x+h by completing the square method.

y + 1 = (-1/4)x^2 - 2x
y +1 = (-1/4)(x^2 + x/2)
y +1 - 1/64 = (-1/4)(x^2 + x/2 + 1/16)
y + 15/16 = (-1/4) (x + 1/4)^2

The vertex would be at point ( -1/4, -15/16)
The focus would be determined as follows:
4p=-1/4 so p=-1/8
focus = (-1/4+(-1/8),-15/16) = (-3/8,-15/16)

Directrix = x = h - p
x = -1/4 - -1/8 = -1/8

Answer 2

Vertex=(-4,3)

Focus=(-4, 2)

Directrix: y=4

Explanation:

Given the equation

`x^2+8x+4y+4=0`

we have to find the vertex, focus, and directrix of parabola.

`x^2+8x+4y+4=0`

`y=(-x^2)/(4)-2x-1`

`h=(-b)/(2a)=(2)/(2((-1)/(4)))=-4`

Now, k can be calculated by putting x=4 and y=k in given equation

`k=-1-(16)/(4)-2(-4)=-1-4+8=3`

The vertex is (h,k) i.e (-4,3)

`y=(-x^2)/(4)-2x-1`

`y=(1)/(4)(-x^2-8x-4)`

`=(1)/(4)(-x^2-8x-16+16-4)`

`y=(1)/(4)(-(x+4)^4+12)`

`y=(-1)/(4)(x+4)^2+3`

which is required vertex form
`4p(y-k)=(x-h)^2`

gives p=-1

Focus=(-4, 3+(-1))=(-4,2)

Now directrix can be calculated as

y=3-p=3-(-1)=4