# Given the equation y^2 + 8x + 8y - 8 = 0 determine: a) The Vertex V b) If the parabola opens up, down, left, or right c) The Focus F

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### 1 Answer

We are given `y^2+8x+8y-8=0` ; we can solve this for x:

`8x=-y^2-8y+8` Complete the square on the right side to get a perfect square trinomial:

`8x=-(y^2+8y-8)`

`8x=-(y^2+8y+16-16-8)`

`8x=-((y+4)^2-24)`

`8x=-(y+4)^2+24`

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

The graph for this is a parabola opening with a horizontal axis, opening to the left.

** The general form for a parabola with a horizontal axis is `x=1/(4p)(y-k)^2+h` with the vertex at (h,k); if p<0 then it opens left, p>0 opens right. p is the focal length, or the distance from the vertex to the focus. **

So `x=-1/8(y+4)^2+3` has a vertex at (3,-4); it opens left; and the focal length is 2. (`1/(4p)=-1/8 ==> p=-2` )

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Vertex (3,-4)

Opens left

Focus at (1,-4)

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The graph:

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Given `(y-k)^2=4p(x-h)` you can find the values of h,k, and p directly from `x=ay^2+by+c` :

`a=1/(4p),b=(-k)/(2p),c=k^2/(4p)+h,h=(4ac-b^2)/(4a),k=(-b)/(2a)`

Here `x=-1/8 y^2-8y+8` so `a=-1/8,b=-1,c=1` and:

`a=1/(4p) ==> p=1/(4a)==>p=1/(4(-1/8))=1/(-1/2)=-2`

`h=(-1/2-1)/(-1/2)=3,k=1/(-1/4)=4`