What is the foci ((x+2)^2)/(100)+((y-5)^2)/(36)=1 ?

The foci ((x+2)^2)/(100)+((y-5)^2)/(36)=1 is (6,5),(-10,5)

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foci ((x+2)^2)/(100)+((y-5)^2)/(36)=1

foci

\frac{( x + 2 ) ^{2}}{100} + \frac{( y - 5 ) ^{2}}{36} = 1

(6, 5), (- 10, 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{100} + \frac{( y - 5 ) ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (- 2, 5), a = 10, b = 6

(- 2 + c, 5), (- 2 - c, 5)

Compute

c :

c = 1 0^{2} - 6^{2} : 8

(- 2 + 8, 5), (- 2 - 8, 5)

Refine

(6, 5), (- 10, 5)

25 x^{2} + 4 y^{2} - 100 x = 0

\frac{x ^{2}}{7} + \frac{y ^{2}}{43} = 1

3 x^{2} + 10 y^{2} = 30

\frac{( x - 1 ) ^{2}}{6} + \frac{( y - 3 ) ^{2}}{2} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{225} = 1

What is the foci ((x+2)^2)/(100)+((y-5)^2)/(36)=1 ?
The foci ((x+2)^2)/(100)+((y-5)^2)/(36)=1 is (6,5),(-10,5)