SOLUTION: we are working on Ellipses and i understand it, but for some reason, i am not coming up with the right answer.
7x^2 + 3y^2 - 28x - 12y = -19
my work:
(7y^2-28x) + (3y^2-12
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-> SOLUTION: we are working on Ellipses and i understand it, but for some reason, i am not coming up with the right answer.
7x^2 + 3y^2 - 28x - 12y = -19
my work:
(7y^2-28x) + (3y^2-12
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Question 538293: we are working on Ellipses and i understand it, but for some reason, i am not coming up with the right answer.
7x^2 + 3y^2 - 28x - 12y = -19
my work:
(7y^2-28x) + (3y^2-12y) = -19
-> 7(x^2 - 4x +(-2)^2) + 3(y^2 - 4y +(-2)^2) = -19 + 28 + 12
-> 7(x-2)^2/21 + 3(y-2)^2/21 = 1
for ellipses aren't the end results for the denominators supposed to be able to be square rooted? i got 3 and 7 which are not able to be square rooted and im confused at what i am doing wrong. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! we are working on Ellipses and i understand it, but for some reason, i am not coming up with the right answer.
7x^2 + 3y^2 - 28x - 12y = -19
my work:
(7y^2-28x) + (3y^2-12y) = -19
-> 7(x^2 - 4x +(-2)^2) + 3(y^2 - 4y +(-2)^2) = -19 + 28 + 12
-> 7(x-2)^2/21 + 3(y-2)^2/21 = 1
for ellipses aren't the end results for the denominators supposed to be able to be square rooted? i got 3 and 7 which are not able to be square rooted and im confused at what i am doing wrong.
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-> 7(x-2)^2/21 + 3(y-2)^2/21 = 1
--> -> (x-2)^2/3 + (y-2)^2/7 = 1
Or
You didn't do anything wrong
