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

Answer:

Vertex=(-4,3)

Focus=(-4, 2)

Directrix: y=4

Step-by-step explanation:

Given the equation

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

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

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

$y=\frac{-x^2}{4}-2x-1$

$h=\frac{-b}{2a}=\frac{2}{2(\frac{-1}{4})}=-4$

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

$k=-1-\frac{16}{4}-2(-4)=-1-4+8=3$

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

$y=\frac{-x^2}{4}-2x-1$

$y=\frac{1}{4}(-x^2-8x-4)$

$=\frac{1}{4}(-x^2-8x-16+16-4)$

$y=\frac{1}{4}(-(x+4)^4+12)$

$y=\frac{-1}{4}(x+4)^2+3$

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