Locate the foci of the ellipse. Show your work.

x^2/36+y^2/11=1

my answer :
General ellipse setup Centre (h,k) is known as :

x^2 = (x-h)^2, then h = 0
y^2 = (x-k)^2, then k = 0
The centre is (0,0)
x^2/36 + y^2/11 = 1
when x= 0 y^2/11 = 1 ; y = 0
When y= 0, x = 0
x^2/36 = 1 ; x = 0
the foci will be along the y=0 axis ... (-c,0) and (c,0)
such that b^2 + c^2 = a^2

Question

Locate the foci of the ellipse. Show your work.

x^2/36+y^2/11=1

my answer : 
General ellipse setup Centre (h,k) is known as :
 
x^2 = (x-h)^2, then h = 0
y^2 = (x-k)^2, then k = 0
 The centre is (0,0)
x^2/36 + y^2/11 = 1
 when x= 0 y^2/11 = 1 ; y = 0
 When y= 0, x = 0 
x^2/36 = 1 ; x = 0
the foci will be along the y=0 axis ... (-c,0) and (c,0) 
such that b^2 + c^2 = a^2

anonymous · Answer

@hartnn please check is it incomplete?

anonymous · Answer

i need to submit this in evening plzz help

hartnn · Answer

yeah, u know how will u get the value of c ?

michele_laino · Answer

please note that:
$$a ^{2}=36, b ^{2}=11$$
so?

anonymous · Answer

11 + c^2 = 36 ; solve for c
C=5

michele_laino · Answer

ok!

hartnn · Answer

yes, c=5 is correct :)

anonymous · Answer

what else?

hartnn · Answer

focus (-5,0) and (5,0)

hartnn · Answer

thats it

michele_laino · Answer

please, substitute now, your value for c, into the coordinates of focuses

anonymous · Answer

confused :/

michele_laino · Answer

you should get as @hartnn  well wrote!

jhannybean · Answer

equation of an ellipse : $\dfrac{x^2}{a^2} +\dfrac{y^2}{b^2}=1$

To find the foci: $a^2 -c^2 =b^2$

michele_laino · Answer

for example first focus, F_1 is:
$$F _{1}=(c,0)=(5,0)$$

michele_laino · Answer

please, do the same with other focus

michele_laino · Answer

$$F _{2}=(-c,0)=...$$

hartnn · Answer

you had foci as 
$(\pm c,0)$
you got c=5
so your foci are 
$(5,0) and (-5,0)$
what else u need ? you're done!