What is the foci (x^2)/(36)+(y^2)/(49)=1 ?

The foci (x^2)/(36)+(y^2)/(49)=1 is (0,sqrt(13)),(0,-sqrt(13))

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foci (x^2)/(36)+(y^2)/(49)=1

foci

\frac{x ^{2}}{36} + \frac{y ^{2}}{49} = 1

(0, 13), (0, - 13)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{36} + \frac{y ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 7, a = 6

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

Compute

c :

c = 7^{2} - 6^{2} : 13

(0, 0 + 13), (0, 0 - 13)

Refine

(0, 13), (0, - 13)

\frac{( x + 7 ) ^{2}}{90} + \frac{( y + 8 ) ^{2}}{84} = 1

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

\frac{x ^{2}}{27} + \frac{y ^{2}}{36} = 1

x^{2} + 4 x + 2 y^{2} + 6 y + 1 = 0

25 x^{2} + 81 y^{2} = 2025

What is the foci (x^2)/(36)+(y^2)/(49)=1 ?
The foci (x^2)/(36)+(y^2)/(49)=1 is (0,sqrt(13)),(0,-sqrt(13))