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

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

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

foci

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

(1, 1), (1, - 5)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (1, - 2), b = 5, a = 4

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

Compute

c :

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

(1, - 2 + 3), (1, - 2 - 3)

Refine

(1, 1), (1, - 5)

9 x^{2} + 25 y^{2} + 90 x = 0

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{9 ^{2}} = 1

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{16} = 1

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

\frac{x ^{2}}{46} + \frac{( y + 8 ) ^{2}}{26} = 1

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