What is the foci ((x^2))/(49)+((y^2))/(25)=1 ?

The foci ((x^2))/(49)+((y^2))/(25)=1 is (2sqrt(6),0),(-2sqrt(6),0)

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foci ((x^2))/(49)+((y^2))/(25)=1

foci

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

(26, 0), (- 26, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = 7, b = 5

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

Compute

c :

c = 7^{2} - 5^{2} : 26

(0 + 26, 0), (0 - 26, 0)

Refine

(26, 0), (- 26, 0)

9 x^{2} + 4 y^{2} + 32 y + 28 = 0

\frac{x ^{2}}{625} + \frac{y ^{2}}{400} = 1

\frac{( y + 4 ) ^{2}}{49} + \frac{( x - 9 ) ^{2}}{25} = 1

1 x^{2} + 3 y^{2} + 9 x + 8 y - 3 = 0

\frac{( x - 5 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{49} = 1

What is the foci ((x^2))/(49)+((y^2))/(25)=1 ?
The foci ((x^2))/(49)+((y^2))/(25)=1 is (2sqrt(6),0),(-2sqrt(6),0)