Finding vertex and focus of parabola given an equation

Question

I am defeated to complete square on the following parabolic equation. Please help. Find the vertex and focal width for the parabola: $$ x^2+6x+8y+1=0 $$

I am hoping to get an equation in this form $$(x−h)^2=4p(y−k)$$

Answer 1

$$x^2 + 6x \color{blue}{+ 9} + 8y + 1 \color{blue}{- 9} = 0$$

$$(x + 3)^2 + 8(y - 1) = 0$$

$$(x + 3)^2 = -8(y - 1) = 4(-2)(y-1)$$

Now, you should be able to "read off" the vertex of the parabola. From there, see if you can find.

With respect to completing the square: you have

$$(x + 3)^2 + 8 y + 1 = 9$$ Subtract $9$ from both sides of the equation. $$\begin{align} (x + 3)^2 + 8y + 1 - 9 = 0 & \iff (x+3)^2 + 8y - 8 = 0 \\ \\ & \iff (x+3)^2 + 8(y - 1) = 0 \end{align}$$

Then subtract $8(y - 1)$ from each side of the equation to get the form you need:

$$(x + 3)^2 = -8(y - 1) = 4(-2)(y - 1)$$

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

So, $$p = -2, \;h = -3,\; k = 1$$

Answer 2

The equation can easily be put in the form you require. Note that $$x^2+6x+8y+1=(x+3)^2+8y-8$$

So you have $$(x+3)^2=4\times-2\times(y-1)$$

Which is the form you wanted.