Question

Find the standard from of x^2 + 4y^2 + 8x - 8y + 16 = 0 then find the foci and the center.

Answer 1

@RadEn can you help??

Answer 2

x^2 + 4y^2 + 8x - 8y + 16 = 0
are u sure the second term (4y^2) ?
it's not a circle equation

Answer 3

yes it says 4y^2 and yea it should be and ellipse

Answer 4

@RadEn are you there?

Answer 5

ohhh... sorry i think it's a circle :)
x^2 + 4y^2 + 8x - 8y + 16 = 0
x^2 + 8x + 16 + 4y^2-8y = 0
(x+4)^2 + 4(y-1)^2 - 4 = 0
(x+4)^2 + 4(y-1)^2 = 4
divided by 4 to both sides, gives
((x+4)^2)/4 + (y-1)^2 = 1
((x-(-4)^2)/2^2 + (y-1)^2/(1)^2 = 1
so, the center is (4,1)
the foci (c^2) = 2^2 - 1^1 = 4 - 1 = 3

Answer 6

ok thanks i'll find the major axis and minor by my self

Answer 7

yw...yes, right u are

Answer 8

excuse me but will the foci be 4 and 3?

Answer 9

why be 4?

Answer 10

so it will be -5.73, -2.27 right?

Answer 11

how come did u get it, show to m

Answer 12

http://www.wolframalpha.com/input/?i=foci+%28%28x%2B4%29%5E2%29%2F4++%2B++%28y-1%29%5E2+%3D+1

Answer 13

that what is says in there

Answer 14

lol, wolfram is the best calculator...
btw, the typo from me above.
the center should be (-4,1)
sorry, my bad :)

Answer 15

okay thanks anyways :)

Answer 16

welcome, nice pic^^

Answer 17

thx