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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

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

4 x^{2} - 8 y^{2} = 3

\frac{y ^{2}}{1600} - \frac{x ^{2}}{5625} = 1

\frac{( x + 2 ) ^{2}}{64} - \frac{( y - 3 ) ^{2}}{36} = 1

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

(\frac{x - 6}{4})^{2} - (\frac{y}{4})^{2} = 1

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