What is the foci (x^2)/(15)+((y-5)^2)/(20)=1 ?

The foci (x^2)/(15)+((y-5)^2)/(20)=1 is (0,5+sqrt(5)),(0,5-sqrt(5))

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foci (x^2)/(15)+((y-5)^2)/(20)=1

foci

\frac{x ^{2}}{15} + \frac{( y - 5 ) ^{2}}{20} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{15} + \frac{( y - 5 ) ^{2}}{20} = 1 :

Ellipse with center

(h, k) = (0, 5), b = 25, a = 15

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

Compute

c :

c = (25)^{2} - 15^{2} : 5

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

x^{2} + 9 y^{2} - 8 x + 18 y - 11 = 0

4 x^{2} + 36 y^{2} - 24 x - 288 y + 611 = 0

\frac{( x - 1 ) ^{2}}{12} + \frac{( y - 2 ) ^{2}}{8} = 1

\frac{( x + 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{16} = 1

1 x^{(2)} + 4 y^{(2)} + 10 x + 9 y - 10 = 0

What is the foci (x^2)/(15)+((y-5)^2)/(20)=1 ?
The foci (x^2)/(15)+((y-5)^2)/(20)=1 is (0,5+sqrt(5)),(0,5-sqrt(5))