What is the foci ((x-4)^2)/(25)+((y+1)^2)/(16)=1 ?

The foci ((x-4)^2)/(25)+((y+1)^2)/(16)=1 is (7,-1),(1,-1)

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foci ((x-4)^2)/(25)+((y+1)^2)/(16)=1

foci

\frac{( x - 4 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

(7, - 1), (1, - 1)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1 :

Ellipse with center

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

(4 + c, - 1), (4 - c, - 1)

Compute

c :

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

(4 + 3, - 1), (4 - 3, - 1)

Refine

(7, - 1), (1, - 1)

\frac{x ^{2}}{9} + \frac{y ^{2}}{49} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 5 ) ^{2}}{4} = 1

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

\frac{( x - 1 ) ^{2}}{81} + \frac{( y - 7 ) ^{2}}{45} = 1

3 x^{2} + 7 y^{2} = 21

What is the foci ((x-4)^2)/(25)+((y+1)^2)/(16)=1 ?
The foci ((x-4)^2)/(25)+((y+1)^2)/(16)=1 is (7,-1),(1,-1)