What is the foci 9x^2+4y^2=25 ?

The foci 9x^2+4y^2=25 is (0,(5sqrt(5))/6),(0,-(5sqrt(5))/6)

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foci 9x^2+4y^2=25

foci

9 x^{2} + 4 y^{2} = 25

(0, \frac{5 5}{6}), (0, - \frac{5 5}{6})

Solution steps

(h, k + c), (h, k - c)

Calculate ellipse properties

9 x^{2} + 4 y^{2} = 25 :

Ellipse with center

(h, k) = (0, 0), b = \frac{5}{2}, a = \frac{5}{3}

(0, 0 + c), (0, 0 - c)

Compute

c :

c = (\frac{5}{2})^{2} - (\frac{5}{3})^{2} : \frac{5 5}{6}

(0, 0 + \frac{5 5}{6}), (0, 0 - \frac{5 5}{6})

Refine

(0, \frac{5 5}{6}), (0, - \frac{5 5}{6})

4 x^{2} + 9 y^{2} + 24 x - 36 y + 36 = 0

4 x^{2} + 16 y^{2} - 16 x - 48 = 0

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

4 x^{2} + 9 y^{2} - 48 x + 72 y = - 144

36 x^{2} + 16 y^{2} - 180 x + 96 y - 207 = 0

What is the foci 9x^2+4y^2=25 ?
The foci 9x^2+4y^2=25 is (0,(5sqrt(5))/6),(0,-(5sqrt(5))/6)