What is the foci (x^2)/(36)+(y^2)/(11)=1 ?

The foci (x^2)/(36)+(y^2)/(11)=1 is (5,0),(-5,0)

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foci (x^2)/(36)+(y^2)/(11)=1

foci

\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1

(5, 0), (- 5, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 6, b = 11

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

Compute

c :

c = 6^{2} - 11^{2} : 5

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

\frac{x ^{2}}{256} + \frac{y ^{2}}{400} = 1

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

25 x^{2} + 16 y^{2} = 400

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

2 x^{2} + y^{2} + 12 x - 6 y + 26 = 0

What is the foci (x^2)/(36)+(y^2)/(11)=1 ?
The foci (x^2)/(36)+(y^2)/(11)=1 is (5,0),(-5,0)