SOLUTION: Given an ellipse: x^2/36+y^2/32=1. Determine the distance between the foci. Please explain.

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Question 1115786: Given an ellipse: x^2/36+y^2/32=1. Determine the distance between the foci. Please explain.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Given an ellipse:
x%5E2%2F36%2By%5E2%2F32=1
This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.
%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1
so, h=0, k=0-> center is at (h,k)= (0,0)
a%5E2=36->a=6
b%5E2=32->b=4sqrt%282%29
Find c, the distance from the center to a focus using following formula:
c=sqrt%28a%5E2-b%5E2%29
c=sqrt%2836-32%29
c=sqrt%284%29
c=2 or c=+-2
Find the foci:
the first focus of an ellipse can be found by adding c to h
.
(h%2Bc,k)=(2,0)
the second foci
(h%2Bc,k)=(0%2B%28-2%29,0)=(-2,0)

the distance between the foci:
d=sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2%2B%28y%5B1%5D-y%5B2%5D%29%5E2%29
d=sqrt%28%282-%28-2%29%29%5E2%2B%280-0%29%5E2%29
d=sqrt%28%282%2B2%29%5E2%29
d=sqrt%284%5E2%29
d=4