What is the foci 9x^2+4y^2+32y+28=0 ?

The foci 9x^2+4y^2+32y+28=0 is (0,-4+sqrt(5)),(0,-4-sqrt(5))

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foci 9x^2+4y^2+32y+28=0

foci

9 x^{2} + 4 y^{2} + 32 y + 28 = 0

(0, - 4 + 5), (0, - 4 - 5)

Solution steps

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

Calculate ellipse properties

9 x^{2} + 4 y^{2} + 32 y + 28 = 0 :

Ellipse with center

(h, k) = (0, - 4), b = 3, a = 2

(0, - 4 + c), (0, - 4 - c)

Compute

c :

c = 3^{2} - 2^{2} : 5

(0, - 4 + 5), (0, - 4 - 5)

\frac{x ^{2}}{625} + \frac{y ^{2}}{400} = 1

\frac{( y + 4 ) ^{2}}{49} + \frac{( x - 9 ) ^{2}}{25} = 1

1 x^{2} + 3 y^{2} + 9 x + 8 y - 3 = 0

\frac{( x - 5 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{49} = 1

\frac{( x + 5 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{25} = 1

What is the foci 9x^2+4y^2+32y+28=0 ?
The foci 9x^2+4y^2+32y+28=0 is (0,-4+sqrt(5)),(0,-4-sqrt(5))