SOLUTION: hello! im doing conic sections and on my homework its asking me to graph {{{x+4+y^2-8y=o}}} im very lost if someone can help me slove this example it would be very helpful.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: hello! im doing conic sections and on my homework its asking me to graph {{{x+4+y^2-8y=o}}} im very lost if someone can help me slove this example it would be very helpful.      Log On


   

Question 599918: hello! im doing conic sections and on my homework its asking me to graph x%2B4%2By%5E2-8y=o im very lost if someone can help me slove this example it would be very helpful.
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
x+4+y^2-8y=0
complete the square
x+4+(y^2-8y+16)=16
x+4+(y-4)^2=16
(y-4)^2=16-x-4
(y-4)^2=-x+12
(y-4)^2=-(x-12)
This is an equation for a parabola that opens leftwards.
Its standard form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of vertex
For given parabola:
vertex: (12,4)
axis of symmetry: y=4
4p=1
p=1/4
focus:(12-p,4)=(12-1/4,4)=(47/4,4) (p-distance left of vertex)
directrix: x=(12+p)=(12+1/4)=49/4 (p-distance right of vertex on axis of symmetry)
see graph below:
y=±(-x+12)^.5+4