Question 616029: If you're given
x^2/25+y^2/16=1
What is the foci?
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Given 36x^2+9y^2-324=0
Find the
center:
vertices:
co-vertices:
foci: I know it's c^2=a^2+b^2 or subtraction?
Sorry, These questions I have trouble with on the review, please help!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! x^2/25+y^2/16=1
What is the foci?
This is an equation of an ellipse with horizontal major axis
Its standard form: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), (h,k)=(x,y) coordinates of the center
For given equation:
center: (0,0)
a^2=25
b^2=16
c^2=a^2-b^2=25-16=9
c=√9=3
foci: (0±c,0)=(0±3,0)=(-3,0) and (3,0)
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Given 36x^2+9y^2-324=0
36x^2+9y^2=324
divide by 324
x^2/9+y^2/36=1
This is an ellipse with vertical major axis
center: (0,0)
..
a^2=36
a=√36=6
vertices:(0,0±a)=(0,0±6)=(0,-6) and (0,6)
..
b^2=9
b=3
co-vertices:(0±b,0)=(0±3,0)=(-3,0) and (3,0)
..
c^2=a^2-b^2=36-9=25
c=√25=5
foci:(0,0±c)=(0,0±5)=(0,-5) and (0,5)
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