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

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the center, vertices and foci of the following ellipse: (x + 3)^2/9 + (y + 1)^2/16 =1      Log On


   

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) About Me  (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:
highlight%28%28x-h%29%5E2%2Fb%5E2%2B%28y-k%29%5E2%2Fa%5E2=1%29
Characteristics:
a>b and b%5E2=a%5E2-c%5E2
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
b%5E2=a%5E2-c%5E2
b%5E2-a%5E2=-c%5E2
-b%5E2%2Ba%5E2=c%5E2
-9%2B16=c%5E2
7=c%5E2
sqrt%287%29=c
Therefore, the center: (h,k)=(-3,-1)
Foci: (h,k+c), (h,k-c)=(-3,-1%2Bsqrt%287%29), (-3,-1-sqrt%287%29)
Vertices: (h,k+a), (h,k-a)=(-3,-1+4),(-3,-1-4)==>(-3,3),(-3,-5)
Happy Calculating!!!!