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

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

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

foci

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

(- 2 + 41, 3), (- 2 - 41, 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 2, 3), a = 4, b = 5

(- 2 + c, 3), (- 2 - c, 3)

Compute

c :

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

(- 2 + 41, 3), (- 2 - 41, 3)

\frac{( x - 8 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

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

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

x^{2} - 9 y^{2} = 9

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

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