What is the foci ((x+3)^2)/(25)-((y-2)^2)/(49)=1 ?

The foci ((x+3)^2)/(25)-((y-2)^2)/(49)=1 is (-3+sqrt(74),2),(-3-sqrt(74),2)

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foci ((x+3)^2)/(25)-((y-2)^2)/(49)=1

foci

\frac{( x + 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{49} = 1

(- 3 + 74, 2), (- 3 - 74, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{49} = 1 :

Right-left Hyperbola with

(h, k) = (- 3, 2), a = 5, b = 7

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

Compute

c :

c = 5^{2} + 7^{2} : 74

(- 3 + 74, 2), (- 3 - 74, 2)

4 y^{2} - x^{2} = 900

- \frac{( x - 2 ) ^{2}}{5} + \frac{( y + 1 ) ^{2}}{9} = 1

\frac{x ^{2}}{64} - \frac{y ^{2}}{32} = 1

25 y^{2} - 81 x^{2} = 2025

\frac{y ^{2}}{4} - \frac{x ^{2}}{16} = 1

What is the foci ((x+3)^2)/(25)-((y-2)^2)/(49)=1 ?
The foci ((x+3)^2)/(25)-((y-2)^2)/(49)=1 is (-3+sqrt(74),2),(-3-sqrt(74),2)