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

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

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

foci

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

(23, 0), (- 23, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

(0 + 23, 0), (0 - 23, 0)

Refine

(23, 0), (- 23, 0)

\frac{x ^{2}}{20} + \frac{y ^{2}}{16} = 1

\frac{y ^{2}}{16} + \frac{x ^{2}}{25} = 1

\frac{1}{4} (x - 5)^{2} + \frac{1}{9} (y + 5)^{2} = 1

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{49} = 1

4 x^{2} + 36 y^{2} = 144

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