What is the foci (y^2)/(25)-((x+3)^2)/(56)=1 ?

The foci (y^2)/(25)-((x+3)^2)/(56)=1 is (-3,9),(-3,-9)

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foci (y^2)/(25)-((x+3)^2)/(56)=1

foci

\frac{y ^{2}}{25} - \frac{( x + 3 ) ^{2}}{56} = 1

(- 3, 9), (- 3, - 9)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{25} - \frac{( x + 3 ) ^{2}}{56} = 1 :

Up-down Hyperbola with

(h, k) = (- 3, 0), a = 5, b = 214

(- 3, 0 + c), (- 3, 0 - c)

Compute

c :

c = 5^{2} + (214)^{2} : 9

(- 3, 0 + 9), (- 3, 0 - 9)

Refine

(- 3, 9), (- 3, - 9)

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

25 x^{2} - 9 y^{2} = 225

\frac{( y - 5 ) ^{2}}{16} - \frac{( x - 4 ) ^{2}}{9} = 1

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

144 y^{2} - 25 x^{2} = 3600

What is the foci (y^2)/(25)-((x+3)^2)/(56)=1 ?
The foci (y^2)/(25)-((x+3)^2)/(56)=1 is (-3,9),(-3,-9)