What is the foci 16x^2-32x+9y^2=128 ?

The foci 16x^2-32x+9y^2=128 is (1,sqrt(7)),(1,-sqrt(7))

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foci 16x^2-32x+9y^2=128

foci

16 x^{2} - 32 x + 9 y^{2} = 128

(1, 7), (1, - 7)

Solution steps

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

Calculate ellipse properties

16 x^{2} - 32 x + 9 y^{2} = 128 :

Ellipse with center

(h, k) = (1, 0), b = 4, a = 3

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

Compute

c :

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

(1, 0 + 7), (1, 0 - 7)

Refine

(1, 7), (1, - 7)

(\frac{x ^{2}}{15}) + (\frac{y ^{2}}{16}) = 1

9 x^{2} - 18 x + 25 y^{2} - 250 y + 409 = 0

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0

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

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

What is the foci 16x^2-32x+9y^2=128 ?
The foci 16x^2-32x+9y^2=128 is (1,sqrt(7)),(1,-sqrt(7))