SOLUTION: Find the center, vertices and foci of the following ellipse: (x + 3)^2/9 + (y + 1)^2/16 =1

Question 74290: Find the center, vertices and foci of the following ellipse:
(x + 3)^2/9 + (y + 1)^2/16 =1
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
Find the center, vertices and foci of the following ellipse:
(x + 3)^2/9 + (y + 1)^2/16 =1
:
The equation for an ellipse whose denominator is largest under the y's is:

Characteristics:
a>b and
Major axis is parallel to the y-axis.
Center:(h,k)
Foci: (h,k+c), (h,k-c)
Vertices: (h,k+a), (h,k-a)
:
Your equation has:
h=-3
k=-1
a^2=16--->a=4
b^2=9---->b=3

Therefore, the center: (h,k)=(-3,-1)
Foci: (h,k+c), (h,k-c)=(-3,), (-3,)
Vertices: (h,k+a), (h,k-a)=(-3,-1+4),(-3,-1-4)==>(-3,3),(-3,-5)
Happy Calculating!!!!
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