Write x² + 8x = -4y - 8 in standard form, then identify the vertex, focus, axis of symmetry, and directrix of the parabola.

Step-by-step solution

1. 1

   Decide which variable to complete the square on. Only $x$ appears squared, so this is a vertical parabola of the family $(x-h)^2 = 4p(y-k)$. Grouping the $x$-terms on one side and leaving $y$ linear on the other is exactly what that form requires.

2. 2

   Complete the square in $x$. Half of the coefficient $8$ is $4$, and $4^2 = 16$, so add $16$ to both sides to keep the equation balanced:

   $x^2 + 8x + 16 = -4y - 8 + 16$

   $(x+4)^2 = -4y + 8$

3. 3

   Factor the right side to expose $4p$ and $k$. Pulling out $-4$ puts the equation into standard form:

   $(x+4)^2 = -4(y-2)$

   Comparing with $(x-h)^2 = 4p(y-k)$ gives $h=-4$, $k=2$ and $4p=-4$, so $p=-1$. The negative $p$ says the parabola opens downward.

4. 4

   Read off the vertex and axis of symmetry. The vertex is $(h,k) = (-4,\,2)$, and the axis of symmetry is the vertical line through it, $x=-4$. Every point of the curve is mirrored across this line.

5. 5

   Locate the focus and directrix using $p$. The focus sits $|p|$ units from the vertex inside the curve, the directrix the same distance on the outside:

   $\text{focus} = (h,\ k+p) = (-4,\ 1), \qquad \text{directrix: } y = k - p = 3$

6. 6

   Check with the definition of a parabola. Any point on the curve must be equidistant from the focus and the directrix. Take $x=0$, which gives $y=-2$: the distance to the focus $(-4,1)$ is $\sqrt{4^2+3^2}=5$, and the distance to the line $y=3$ is $|-2-3| = 5$. The two agree, confirming the focus and directrix.