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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

\frac{x ^{2}}{36} + \frac{y ^{2}}{4} = 1

f (x) = x^{2} - 10 x + 24

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

f (x) = x^{2} - 8 x + 12

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