What is the foci ((x-3)^2)/(49)+((y+5)^2)/(25)=1 ?

The foci ((x-3)^2)/(49)+((y+5)^2)/(25)=1 is (3+2sqrt(6),-5),(3-2sqrt(6),-5)

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foci ((x-3)^2)/(49)+((y+5)^2)/(25)=1

foci

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

(3 + 26, - 5), (3 - 26, - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (3, - 5), a = 7, b = 5

(3 + c, - 5), (3 - c, - 5)

Compute

c :

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

(3 + 26, - 5), (3 - 26, - 5)

\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1

9 x^{2} + 16 y^{2} = 576

3 x^{2} + y^{2} + 2 x - 6 y = 3

9 x^{2} + 25 y^{2} + 18 x + 150 y + 9 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{49} = 1

What is the foci ((x-3)^2)/(49)+((y+5)^2)/(25)=1 ?
The foci ((x-3)^2)/(49)+((y+5)^2)/(25)=1 is (3+2sqrt(6),-5),(3-2sqrt(6),-5)