What is the foci ((x+1)^2)/9-((y-4)^2)/(49)=1 ?

The foci ((x+1)^2)/9-((y-4)^2)/(49)=1 is (-1+sqrt(58),4),(-1-sqrt(58),4)

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foci ((x+1)^2)/9-((y-4)^2)/(49)=1

foci

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 4 ) ^{2}}{49} = 1

(- 1 + 58, 4), (- 1 - 58, 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 4 ) ^{2}}{49} = 1 :

Right-left Hyperbola with

(h, k) = (- 1, 4), a = 3, b = 7

(- 1 + c, 4), (- 1 - c, 4)

Compute

c :

c = 3^{2} + 7^{2} : 58

(- 1 + 58, 4), (- 1 - 58, 4)

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

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

x^{2} - 2 x - 3 = y^{2} + 2 y + 3

6 x^{2} - 5 y^{2} - 24 x + 10 y - 11 = 0

4 x^{2} - 24 x - 25 y^{2} + 250 y - 489 = 0

What is the foci ((x+1)^2)/9-((y-4)^2)/(49)=1 ?
The foci ((x+1)^2)/9-((y-4)^2)/(49)=1 is (-1+sqrt(58),4),(-1-sqrt(58),4)