maths
posted by salome on .
Given the equation of an ellipse as
5x^2+3y^215=0
find the centre and foci

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 yaxis)