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

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

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

foci

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

(2 + 26, - 1), (2 - 26, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, - 1), a = 7, b = 5

(2 + c, - 1), (2 - c, - 1)

Compute

c :

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

(2 + 26, - 1), (2 - 26, - 1)

25 x^{2} + 9 y^{2} + 100 x - 125 = 0

4 x^{2} + 25 y^{2} - 40 x + 150 y + 225 = 0

\frac{( x - 8 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{5 ^{2}} = 1

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

25 x^{2} + 64 y^{2} = 1600

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