What is the foci (x^2)/(64)+(y^2)/(49)=1 ?

The foci (x^2)/(64)+(y^2)/(49)=1 is (sqrt(15),0),(-sqrt(15),0)

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foci (x^2)/(64)+(y^2)/(49)=1

foci

\frac{x ^{2}}{64} + \frac{y ^{2}}{49} = 1

(15, 0), (- 15, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{64} + \frac{y ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 8, b = 7

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

Compute

c :

c = 8^{2} - 7^{2} : 15

(0 + 15, 0), (0 - 15, 0)

Refine

(15, 0), (- 15, 0)

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0

16 x^{2} - 64 x + 9 y^{2} = 80

4 x^{2} + 9 y^{2} - 8 x - 32 = 0

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

\frac{( x + 3 ) ^{2}}{21} + \frac{( y - 6 ) ^{2}}{25} = 1

What is the foci (x^2)/(64)+(y^2)/(49)=1 ?
The foci (x^2)/(64)+(y^2)/(49)=1 is (sqrt(15),0),(-sqrt(15),0)