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Question 198908: Hi my name is Liz I need help with this question
What are the coordinates of the foci of the graph of the equation 4x2+16y2=16?
Answer by J2R2R(94)   (Show Source):
You can put this solution on YOUR website! 4x^2 + 16y^2 = 16
is
x^2/4 + y^2/1 = 1
or
x^2/2^2 + y^2/1^2 = 1
This is
X^2/a^2 + y^2/b^2 = 1 where a=2 and b=1.
b^2 = a^2(1-e^2) for the eccentricity e and the foci +/- ae
1 = 4 (1-e^2); ¼ = 1-e^2 giving e = (3^0.5)/2
ae = 3^0.5
Therefore the foci are (-3^0.5, 0) and (3^0.5, 0) which are approximately (-1.73205, 0) and (1.73205, 0)
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