What is the foci 16x^2+64x+4y^2-8y+4=0 ?

The foci 16x^2+64x+4y^2-8y+4=0 is (-2,1+2sqrt(3)),(-2,1-2sqrt(3))

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foci 16x^2+64x+4y^2-8y+4=0

foci

16 x^{2} + 64 x + 4 y^{2} - 8 y + 4 = 0

(- 2, 1 + 23), (- 2, 1 - 23)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 64 x + 4 y^{2} - 8 y + 4 = 0 :

Ellipse with center

(h, k) = (- 2, 1), b = 4, a = 2

(- 2, 1 + c), (- 2, 1 - c)

Compute

c :

c = 4^{2} - 2^{2} : 23

(- 2, 1 + 23), (- 2, 1 - 23)

\frac{x ^{2}}{17} + \frac{y ^{2}}{35} = 1

64 x^{2} + 81 y^{2} = 5184

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

4 x^{2} + y^{2} + 40 x + 6 y + 95 = 0

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

What is the foci 16x^2+64x+4y^2-8y+4=0 ?
The foci 16x^2+64x+4y^2-8y+4=0 is (-2,1+2sqrt(3)),(-2,1-2sqrt(3))