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)

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{169} = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{y ^{2}}{9} = 1

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

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

25 x^{2} - 200 x + y^{2} + 4 y + 379 = 0

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)