What is the foci ((x+2)^2)/(64)-((y+4)^2)/(81)=1 ?

The foci ((x+2)^2)/(64)-((y+4)^2)/(81)=1 is (-2+sqrt(145),-4),(-2-sqrt(145),-4)

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foci ((x+2)^2)/(64)-((y+4)^2)/(81)=1

foci

\frac{( x + 2 ) ^{2}}{64} - \frac{( y + 4 ) ^{2}}{81} = 1

(- 2 + 145, - 4), (- 2 - 145, - 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 2 ) ^{2}}{64} - \frac{( y + 4 ) ^{2}}{81} = 1 :

Right-left Hyperbola with

(h, k) = (- 2, - 4), a = 8, b = 9

(- 2 + c, - 4), (- 2 - c, - 4)

Compute

c :

c = 8^{2} + 9^{2} : 145

(- 2 + 145, - 4), (- 2 - 145, - 4)

4 x^{2} - 9 y^{2} - 16 x + 18 y - 7 = 0

x^{2} - 3 y^{2} + 2 x + 6 y - 1 = 0

- 4 x^{2} + 25 y^{2} - 16 x - 116 = 0

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

- \frac{( x + 1 ) ^{2}}{25} + \frac{( y + 5 ) ^{2}}{16} = 1

What is the foci ((x+2)^2)/(64)-((y+4)^2)/(81)=1 ?
The foci ((x+2)^2)/(64)-((y+4)^2)/(81)=1 is (-2+sqrt(145),-4),(-2-sqrt(145),-4)