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
$$(x−h)^2=4p(y−k)$$
[Math] Finding vertex and focus of parabola given an equation
conic sectionsquadratics
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Best Answer
$$x^2 + 6x \color{blue}{+ 9} + 8y + 1 \color{blue}{- 9} = 0$$
$$(x + 3)^2 + 8(y - 1) = 0$$
$$(x + 3)^2 = -8(y - 1) = 4(-2)(y-1)$$
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, $$p = -2, \;h = -3,\; k = 1$$