Question

Find the foci for the ellipse of (x+8)^2/16+(y+2)^2/4=1

Answer 1

first determine your a and b:
a is always greater than b

a^2 = 16
b^2 = 4

--> c^2 = a^2 - b^2
--> c^2 = 12
--> c = 2sqrt3
this is distance the foci are from center
center: (-8,-2)
major axis is horizontal because a^2 is under "x" term

foci: (-8 +-2sqrt3, -2)

Answer 2

what other way can you writhe the answer out cause im taking an class online and its not letting me put it in as that

Answer 3

and thank for exlplaining it to me

Answer 4

thanks*

Answer 5

well there are 2 focus points in an ellipse right. so im guessing it will ask for 2 different points
i just wrote it as plus/minus, so split it into 2 points....one with -8+2sqrt3 and -8 -2sqrt3