What is the foci ((x-3)^2)/(25)+((y+1)^2)/(16)=1 ?

The foci ((x-3)^2)/(25)+((y+1)^2)/(16)=1 is (6,-1),(0,-1)

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foci ((x-3)^2)/(25)+((y+1)^2)/(16)=1

foci

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

(6, - 1), (0, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (3, - 1), a = 5, b = 4

(3 + c, - 1), (3 - c, - 1)

Compute

c :

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

(3 + 3, - 1), (3 - 3, - 1)

Refine

(6, - 1), (0, - 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

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

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

What is the foci ((x-3)^2)/(25)+((y+1)^2)/(16)=1 ?
The foci ((x-3)^2)/(25)+((y+1)^2)/(16)=1 is (6,-1),(0,-1)