Question

find the foci of the ellispe with the equation x^2/16+y^2/25=1

a.(+3,0) b.(0,+3) c.(+/41,0) or d. (0,+/41)

Answer 1

64?

Answer 2

How'd you get 64?

Answer 3

i didnt their was a pic and it said 64, and i said 64? as in how is it 64 its not even an option lol

Answer 4

To find the foci, you use the following equation..

c^2 = a^2 - b^2

Answer 5

so you have
$$
\cfrac{x^2}{16}+\cfrac{y^2}{25}=1 \implies
\cfrac{(x-0)^2}{4^2}+\cfrac{(y-0)^2}{5^2}=1\\
\text{the distance from the center to any of the focus is "c"}\\
\color{blue}{c= \sqrt{a^2-b^2}}
$$
notice your center is (0, 0)

Answer 6

?

Answer 7

so, what would you get for the "c" distance?

Answer 8

i have no idea how to do this stuff thats why i came here for the answer, none of my friend know how either

Answer 9

well, we're under the assumption you've covered this material already, otherwise then this wouldn't apply to you

Answer 10

http://www.youtube.com/watch?v=lvAYFUIEpFI

Answer 11

c^2 = 25 - 16
c^2 = 9
c = 3

Answer 12

So eliminate c and d

Answer 13

thank u

Answer 14

the alebra class i was in we only did have the book and next year we do the next half so i havent got to this yet..