How do you find the coordinates of the center, foci, lengths of the major and minor axes given #4x^2+8y^2=32#?

1 Answer
Narad T.
Oct 29, 2016

The center is #(0,0)#
The foci are #(2,0)# and #(-2,0)#
Length of major axis is #=2sqrt8#
Length of minor axis is#=4#

Explanation:

#4x^2+8y^2=32#
Dividing by 32, we get
#x^2/8+y^2/4=1#
Comparing this to the equation of the ellipse
#(x-x_0)^2/a^2+(y-y_0^2)/b^2=1#

The center is #(x_0,y_0)=(0,0)#
To determine the foci, we calculate #c=sqrt(a^2-b^2)=sqrt(8-4)=sqrt4=2#
Therefore the foci are #(2,0)# and #(-2,0)#
Length of half the major axis is #a=sqrt8#
and the length of half the minor axis is #b=2#
graph{4x^2+8y^2=32 [-6.244, 6.243, -3.12, 3.123]}

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