What is the foci ((x-5)^2)/(25)+((y+7)^2)/(16)=1 ?

The foci ((x-5)^2)/(25)+((y+7)^2)/(16)=1 is (8,-7),(2,-7)

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foci ((x-5)^2)/(25)+((y+7)^2)/(16)=1

foci

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

(8, - 7), (2, - 7)

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1 :

Ellipse with center

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

(5 + c, - 7), (5 - c, - 7)

Compute

c :

c = 5^{2} - 4^{2} : 3

(5 + 3, - 7), (5 - 3, - 7)

Refine

(8, - 7), (2, - 7)

25 x^{2} + 16 y^{2} = 400

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

2 x^{2} + y^{2} + 12 x - 6 y + 26 = 0

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1

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

What is the foci ((x-5)^2)/(25)+((y+7)^2)/(16)=1 ?
The foci ((x-5)^2)/(25)+((y+7)^2)/(16)=1 is (8,-7),(2,-7)