What is the foci ((x-1)^2)/9+((y+4)^2)/(25)=1 ?

The foci ((x-1)^2)/9+((y+4)^2)/(25)=1 is (1,0),(1,-8)

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foci ((x-1)^2)/9+((y+4)^2)/(25)=1

foci

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

(1, 0), (1, - 8)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1 :

Ellipse with center

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

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

Compute

c :

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

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

Refine

(1, 0), (1, - 8)

\frac{x ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{81} = 1

16 xx + 25 yy = 400

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

16 x^{2} + 25 y^{2} - 64 x - 336 = 0

What is the foci ((x-1)^2)/9+((y+4)^2)/(25)=1 ?
The foci ((x-1)^2)/9+((y+4)^2)/(25)=1 is (1,0),(1,-8)