What is the foci 16x^2+25y^2+32x-384=0 ?

The foci 16x^2+25y^2+32x-384=0 is (2,0),(-4,0)

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foci 16x^2+25y^2+32x-384=0

foci

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

(2, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} + 32 x - 384 = 0 :

Ellipse with center

(h, k) = (- 1, 0), a = 5, b = 4

(- 1 + c, 0), (- 1 - c, 0)

Compute

c :

c = 5^{2} - 4^{2} : 3

(- 1 + 3, 0), (- 1 - 3, 0)

Refine

(2, 0), (- 4, 0)

3 x^{2} + 7 y^{2} - 12 x - 28 y + 19 = 0

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

What is the foci 16x^2+25y^2+32x-384=0 ?
The foci 16x^2+25y^2+32x-384=0 is (2,0),(-4,0)