What is the foci (x^2)/(64)+(y^2)/(16)=1 ?

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

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foci (x^2)/(64)+(y^2)/(16)=1

foci

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

(43, 0), (- 43, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 8, b = 4

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

Compute

c :

c = 8^{2} - 4^{2} : 43

(0 + 43, 0), (0 - 43, 0)

Refine

(43, 0), (- 43, 0)

\frac{y ^{2}}{4} - \frac{x ^{2}}{9} = 1

2 y^{2} - 16 x = 0

\frac{x ^{2}}{100} + \frac{y ^{2}}{36} = 1

16 x^{2} - 9 y^{2} = 144

x^{2} - y^{2} = 16

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