What is the foci (x^2)/(64)+(y^2)/(36)=1 ?

The foci (x^2)/(64)+(y^2)/(36)=1 is (2sqrt(7),0),(-2sqrt(7),0)

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foci (x^2)/(64)+(y^2)/(36)=1

foci

\frac{x ^{2}}{64} + \frac{y ^{2}}{36} = 1

(27, 0), (- 27, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{64} + \frac{y ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 8, b = 6

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

Compute

c :

c = 8^{2} - 6^{2} : 27

(0 + 27, 0), (0 - 27, 0)

Refine

(27, 0), (- 27, 0)

x^{2} + 4 y^{2} = 16

(x - 6)^{2} + (y + 5)^{2} = 16

4 x^{2} + 4 y^{2} = 25

x^{2} + y^{2} - 12 x - 10 y = - 12

x^{2} + y^{2} - 6 x = 7

What is the foci (x^2)/(64)+(y^2)/(36)=1 ?
The foci (x^2)/(64)+(y^2)/(36)=1 is (2sqrt(7),0),(-2sqrt(7),0)