SOLUTION: Graph, find center, vertices, co-vertices and foci. 9x^2+4y=36 I worked it out to: center(0,0) a=3 b=2 c=5 vertices(0,+-3) co-vertices(+-2,0) foci(0,+-5) eccentricity=5/3

Question 190916This question is from textbook saxon algebra 2
: Graph, find center, vertices, co-vertices and foci.
9x^2+4y=36
I worked it out to:
center(0,0)
a=3 b=2 c=5
vertices(0,+-3)
co-vertices(+-2,0)
foci(0,+-5)
eccentricity=5/3
Did I do this right? When I graph it, the foci is outside the circle. Is that right? This question is from textbook saxon algebra 2

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Graph, find center, vertices, co-vertices and foci.
9x^2+4y=36
y = (-9/4)x^2 + 9
---------
It's not a circle ; it's a parabola
There is no center and no co-vertices.
---------
Vertex form by squaring the x-terms:
9(x-0)^2 = -4(y-9)
(x-0)^2 = (-4/9)(y-9)
h = 0; k = 9
4p = -4/9
p = -1/9
---------------------------
Vertex: (0 , 9)
Focus (0 , 80/9)
======================
Cheers,
Stan H.
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