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Question

Given the equation of an ellipse as
5x^2+3y^2-15=0
find the centre and foci
13 years ago

Answers

Reiny
change it to the standard form of an ellipse.

5x^2 + 3y^2 = 15
divide each term by 15

x^2 /3 + y^2 /5 = 1

centre is (0,0)
a = √3 , b=√5
c^2 + √3^2 = √5^2
c^2 + 3 = 5
c^2 = 2
c = ± √2

foci : (0,√2) and (0,-√2) , (on the y-axis)
13 years ago

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