Question

What are the foci of (x+3)^2/169 + (y+5)^2/144=1?

(–8, 5) and (2, 5)
(–2, 5) and (8, 5)
(–8, –5) and (2, –5)
(–2, –5) and (8, –5)

Answer 1

a^2=169
b^2=144
c^2=a^2-b^2=25
so c=5.

the foci is \(\pm 5 \) units left/right of the center of the ellipse.
can you identify the center to get the coordinates of your foci?

Answer 2

so which answer would it be?

Answer 3

can you identify the center of the ellipse?

Answer 4

(-3,-5)?

Answer 5

so the foci is
(-3+5, -5) and (-3-5, -5)

Answer 6

thank you!

Answer 7

yw...:)

Answer 8

do you know this one?
What is the focus of the parabola x =1/6 (y + 3)2 – 6 ?

Answer 9

yes...
get that equation into this form: 4a(x-h) = (y-k)^2

your vertex is (h,k) and the focus is "a" units away from the vertex...
so you'll need to find "a"

Answer 10

I`m not sure what the vertex is..

Answer 11

ok... let's start with the equation... add 6 to both sides... write down what you have...

Answer 12

oh i just realized that its supposed to be x=1/16 not just 6!