Posted by **salome** on Sunday, March 11, 2012 at 8:37am.

Given the equation of an ellipse as

5x^2+3y^2-15=0

find the centre and foci

- maths -
**Reiny**, Sunday, March 11, 2012 at 8:44am
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)

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