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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

\frac{( x ) ^{2}}{49} + \frac{( y ) ^{2}}{25} = 1

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 7 ) ^{2}}{36} = 1

\frac{( x - 1 ) ^{2}}{3} + \frac{( y - 2 ) ^{2}}{4} = 1

x^{2} + 4 y^{2} = 36

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