What is the foci 9x^2+16y^2+72x+128y+256=0 ?

The foci 9x^2+16y^2+72x+128y+256=0 is (-4+sqrt(7),-4),(-4-sqrt(7),-4)

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foci 9x^2+16y^2+72x+128y+256=0

foci

9 x^{2} + 16 y^{2} + 72 x + 128 y + 256 = 0

(- 4 + 7, - 4), (- 4 - 7, - 4)

Solution steps

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

Calculate ellipse properties

9 x^{2} + 16 y^{2} + 72 x + 128 y + 256 = 0 :

Ellipse with center

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

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

Compute

c :

c = 4^{2} - 3^{2} : 7

(- 4 + 7, - 4), (- 4 - 7, - 4)

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

\frac{x ^{2}}{15} + \frac{y ^{2}}{64} = 1

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{25} = 1

What is the foci 9x^2+16y^2+72x+128y+256=0 ?
The foci 9x^2+16y^2+72x+128y+256=0 is (-4+sqrt(7),-4),(-4-sqrt(7),-4)