What is the foci (x^2)/(36)+(y^2)/(25)=1 ?

The foci (x^2)/(36)+(y^2)/(25)=1 is (sqrt(11),0),(-sqrt(11),0)

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foci (x^2)/(36)+(y^2)/(25)=1

foci

\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1

(11, 0), (- 11, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 6, b = 5

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

Compute

c :

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

(0 + 11, 0), (0 - 11, 0)

Refine

(11, 0), (- 11, 0)

x^{2} + y^{2} - 2 y = 0

9 x^{2} - 9 y^{2} + 18 x + 36 y = 63

y^{2} = 28 x

\frac{x ^{2}}{64} - \frac{y ^{2}}{9} = 1

\frac{y ^{2}}{16} - \frac{x ^{2}}{25} = 1

What is the foci (x^2)/(36)+(y^2)/(25)=1 ?
The foci (x^2)/(36)+(y^2)/(25)=1 is (sqrt(11),0),(-sqrt(11),0)