SOLUTION: I am having a hard time graphing this type of problem. I have the following: (x + 3)^2 / 16 + (y - 2)^2 /36 = 1 I have the following (x + 3)^2 / 4^2 + (y

SOLUTION: I am having a hard time graphing this type of problem. I have the following: (x + 3)^2 / 16 + (y - 2)^2 /36 = 1 I have the following (x + 3)^2 / 4^2 + (y - 2 )^2 / 6^2 =

Question 516813: I am having a hard time graphing this type of problem. I have the following:
(x + 3)^2 / 16 + (y - 2)^2 /36 = 1
I have the following
(x + 3)^2 / 4^2 + (y - 2 )^2 / 6^2 = 1
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1. (x + 3)^2 / 16 + (y - 2)^2 /36 = 1
2. (x + 3)^2 / 4^2 + (y - 2 )^2 / 6^2 = 1
**
Both of these equations are ellipses with vertical major axis of the standard form:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, with (h,k) being the (x,y) coordinates of the center.
..
(note: If the denominator under the x-term was greater than the denominator under the y-term, the ellipse would have a horizontal major axis, and the standard form would be:
(x-h)^2/a^2+(y-k)^2/b^2=1)
..
For equation 1.
center: (-3,2)
a^2=36
a=√36=6
length of vertical major axis=2a=12
vertices(end-points of major axis): (-3,2�a)=(-3,2�6)=(-3,8) and (-3,-4)
b^2=16
b=√16=4
length of minor axis=2b=8
end-points of minor axis: (-3�b,2)=(-3�4,2)=(-7,2) and (1,2)
Foci
c^2=a^2-b^2=36-16=20
c=√20≈4.47
foci=(-3,2�c)=(-3,2�4.47)=(-3,6.47) and (-3,-2.27)
You now have all the coordinates you need to graph equation 1.
I will leave it up to you to use the same procedure to find the coordinates you need to graph equation 2.
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