SOLUTION: 11. What is the center of the ellipse whose equation is (x^2)/25 + (y^2)/4 = 1? A (25,1) B (5,1) C (4,25) D (4,1) E (0,0) F (5,2) (x-h)^2 / a^2 + (y-k)^2 / b^2 =

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 11. What is the center of the ellipse whose equation is (x^2)/25 + (y^2)/4 = 1? A (25,1) B (5,1) C (4,25) D (4,1) E (0,0) F (5,2) (x-h)^2 / a^2 + (y-k)^2 / b^2 =       Log On


   

Question 162020: 11. What is the center of the ellipse whose equation is (x^2)/25 + (y^2)/4 = 1?
A (25,1)
B (5,1)
C (4,25)
D (4,1)
E (0,0)
F (5,2)

(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1 I tried to used this equation to help, but im really stuck. I appericaite any help you can offer. Thank you!
Found 2 solutions by scott8148, jim_thompson5910:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
from the given equation, h and k are both zero __ a and b have nothing to do with the center

E looks good...

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You have the right idea, but it might help to go even further

%28x%5E2%29%2F25+%2B+%28y%5E2%29%2F4+=+1 Start with the given equation


%28x-0%29%5E2%2F25+%2B+%28y-0%29%5E2%2F4+=+1 Replace x with x-0. Replace y with y-0. This is possible since adding/subtracting zero doesn't affect the equation


%28x-0%29%5E2%2F5%5E2+%2B+%28y-0%29%5E2%2F2%5E2+=+1 Rewrite 25 as 5%5E2. Rewrite 4 as 2%5E2.



Notice how the equation above fits the form %28x-h%29%5E2+%2F+a%5E2+%2B+%28y-k%29%5E2+%2F+b%5E2+=+1 where in this case h=0, a=5, k=0, and b=2.

Remember the center is simply (h,k). So this means that the center is (0,0). So the answer is E


Note: If you are really stuck, you can solve for "y" and graph. Once you do that, you'll see that the center is at (0,0)