# Finding vertex and focus of parabola given an equation

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 $$(x−h)^2=4p(y−k)$$

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I am hoping to get an equation in this form $$(x-h)^2 = 4p(y-k)$$. problem is, when I try complete the square method on the equation, I only find $$(x+3)^2 + 8y + 1 =9$$ –  Sylvester Sep 15 '13 at 22:31
Subtract $9$ from both sides of the equation. $8y + 1 - 9 = 8y - 8 = 8(y - 1)$. Then subtract $8(y - 1)$ from each side of the equation to get the form you need. –  amWhy Sep 15 '13 at 22:39

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

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

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

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

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Fantastic. I just couldn't see that. Thanks. –  Sylvester Sep 15 '13 at 23:24
am just wondering in you solutions: why is k not -2. –  Sylvester Sep 15 '13 at 23:46
Recall that $p$ appears as a factor preceding the $(y - k)$ term: $4 \cdot -2\cdot (y - k)$ –  amWhy Sep 15 '13 at 23:49
am sorry @amWhy. I actually meant why is k not -1 since (y-k) = (y-1). We just pluck out figures. –  Sylvester Sep 16 '13 at 0:03
@amWhy: Also needs a TU to go with the check +1 –  Amzoti Sep 16 '13 at 0:27

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.

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Hi @Mark-bennet, Which equation is mistaken? I thought the standard parabola is expressed as $$x^2 = 4py$$ or expanded as above with vertex values (h,k)? Please clarify. –  Sylvester Sep 15 '13 at 22:45
@Sylvester apologies - read 9 as 1 - adjusted in answer –  Mark Bennet Sep 15 '13 at 22:47