SOLUTION: i need help finding the center, vertices, and foci of this equation 9x^2+36(y-2)^2-324=0

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Question 765278: i need help finding the center, vertices, and foci of this equation 9x^2+36(y-2)^2-324=0
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
9x%5E2%2B36%28y-2%29%5E2-324=0

9x%5E2%2B36%28y-2%29%5E2=324
9x%5E2%2F324%2B36%28y-2%29%5E2%2F324=324%2F324
cross%289%29x%5E2%2Fcross%28324%2936%2Bcross%2836%29%28y-2%29%5E2%2Fcross%28324%299=1
x%5E2%2F36%2B%28y-2%29%5E2%2F9=1 or
%28x-0%29%5E2%2F36%2B%28y-2%29%5E2%2F9=1
this is an ellipse with h=0 and k=2; so, the center is at
(0,2)
we also know that semi-major axis a=6 and semi-minor axis b=3; so, we have an ellipse with horizontal major axis
vertices:
(h-a,k) =>(0-6,2)=>(-6,2)
(h%2Ba,k)=>(0%2B6,2)=>(6,2)
c0-vertices:
(k,k-b) =>(2,2-3)=>(2,-1)
(k,k%2Bb)=>(2,2%2B3)=>(2,5)
foci:
The formula generally associated with the focus of an ellipse is
+c%5E2=+a%5E2+-+b%5E2
so, b=3 and a=6
+c%5E2=+6%5E2+-+3%5E2
+c%5E2=36+-+9
+c%5E2=27
+c=sqrt%2827%29
+c=5.19 and +c=-5.19
so, foci is at (5.19,2) and (-5.19,2)