What is the foci 16x^2+25y^2+64x-336=0 ?

The foci 16x^2+25y^2+64x-336=0 is (1,0),(-5,0)

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foci 16x^2+25y^2+64x-336=0

foci

16 x^{2} + 25 y^{2} + 64 x - 336 = 0

(1, 0), (- 5, 0)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} + 64 x - 336 = 0 :

Ellipse with center

(h, k) = (- 2, 0), a = 5, b = 4

(- 2 + c, 0), (- 2 - c, 0)

Compute

c :

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

(- 2 + 3, 0), (- 2 - 3, 0)

Refine

(1, 0), (- 5, 0)

\frac{( x - 7 ) ^{2}}{81} + \frac{( y - 6 ) ^{2}}{64} = 1

\frac{( X + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

6 x^{2} + 10 y^{2} = 60

4 x^{2} + 9 y^{2} + 24 x - 90 y = - 225

\frac{x ^{2}}{36} + \frac{y ^{2}}{20} = 1

What is the foci 16x^2+25y^2+64x-336=0 ?
The foci 16x^2+25y^2+64x-336=0 is (1,0),(-5,0)