What is the foci ((y+4)^2)/(49)+((x-9)^2)/(25)=1 ?

The foci ((y+4)^2)/(49)+((x-9)^2)/(25)=1 is (9,-4+2sqrt(6)),(9,-4-2sqrt(6))

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foci ((y+4)^2)/(49)+((x-9)^2)/(25)=1

foci

\frac{( y + 4 ) ^{2}}{49} + \frac{( x - 9 ) ^{2}}{25} = 1

(9, - 4 + 26), (9, - 4 - 26)

Solution steps

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

Calculate ellipse properties

\frac{( y + 4 ) ^{2}}{49} + \frac{( x - 9 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (9, - 4), b = 7, a = 5

(9, - 4 + c), (9, - 4 - c)

Compute

c :

c = 7^{2} - 5^{2} : 26

(9, - 4 + 26), (9, - 4 - 26)

1 x^{2} + 3 y^{2} + 9 x + 8 y - 3 = 0

\frac{( x - 5 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{49} = 1

\frac{( x + 5 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{25} = 1

4 x^{2} + 9 y^{2} + 16 x + 18 y - 11 = 0

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

What is the foci ((y+4)^2)/(49)+((x-9)^2)/(25)=1 ?
The foci ((y+4)^2)/(49)+((x-9)^2)/(25)=1 is (9,-4+2sqrt(6)),(9,-4-2sqrt(6))