What is the foci x^2+10x+9y^2+16=0 ?

The foci x^2+10x+9y^2+16=0 is (-5+2sqrt(2),0),(-5-2sqrt(2),0)

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foci x^2+10x+9y^2+16=0

foci

x^{2} + 10 x + 9 y^{2} + 16 = 0

(- 5 + 22, 0), (- 5 - 22, 0)

Solution steps

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

Calculate ellipse properties

x^{2} + 10 x + 9 y^{2} + 16 = 0 :

Ellipse with center

(h, k) = (- 5, 0), a = 3, b = 1

(- 5 + c, 0), (- 5 - c, 0)

Compute

c :

c = 3^{2} - 1^{2} : 22

(- 5 + 22, 0), (- 5 - 22, 0)

9 x^{2} + y^{2} + 18 x + 4 y + 4 = 0

5 x^{2} + 2 y^{2} = 50

4 x^{2} + 49 y^{2} = 196

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

16 x^{2} - 96 x + 4 y^{2} = - 80

What is the foci x^2+10x+9y^2+16=0 ?
The foci x^2+10x+9y^2+16=0 is (-5+2sqrt(2),0),(-5-2sqrt(2),0)