What is the foci ((y-3)^2)/(16)-((x+4)^2)/(25)=1 ?

The foci ((y-3)^2)/(16)-((x+4)^2)/(25)=1 is (-4,3+sqrt(41)),(-4,3-sqrt(41))

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foci ((y-3)^2)/(16)-((x+4)^2)/(25)=1

foci

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

(- 4, 3 + 41), (- 4, 3 - 41)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1 :

Up-down Hyperbola with

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

(- 4, 3 + c), (- 4, 3 - c)

Compute

c :

c = 4^{2} + 5^{2} : 41

(- 4, 3 + 41), (- 4, 3 - 41)

16 x^{2} - y^{2} - 64 x - 2 y + 47 = 0

\frac{( x - 1 ) ^{2}}{16} - \frac{( y - 7 ) ^{2}}{20} = 1

y^{2} - 9 x^{2} - 8 y - 54 x - 81 = 0

\frac{( x - 3 ) ^{2}}{8 ^{2}} - \frac{( y - 2 ) ^{2}}{8 ^{2}} = 1

\frac{x ^{2}}{49} - \frac{y ^{2}}{81} = 1

What is the foci ((y-3)^2)/(16)-((x+4)^2)/(25)=1 ?
The foci ((y-3)^2)/(16)-((x+4)^2)/(25)=1 is (-4,3+sqrt(41)),(-4,3-sqrt(41))