What is the foci ((x+2)^2)/9-((y-1)^2)/(169)=1 ?

The foci ((x+2)^2)/9-((y-1)^2)/(169)=1 is (-2+sqrt(178),1),(-2-sqrt(178),1)

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foci ((x+2)^2)/9-((y-1)^2)/(169)=1

foci

\frac{( x + 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{169} = 1

(- 2 + 178, 1), (- 2 - 178, 1)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{169} = 1 :

Right-left Hyperbola with

(h, k) = (- 2, 1), a = 3, b = 13

(- 2 + c, 1), (- 2 - c, 1)

Compute

c :

c = 3^{2} + 1 3^{2} : 178

(- 2 + 178, 1), (- 2 - 178, 1)

y^{2} - 49 x^{2} - 14 y - 490 x - 1125 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{4} = 1

4 y^{2} - x^{2} + 6 x + 8 y + 4 = 0

- 9 x^{2} + y^{2} + 6 y - 27 = 0

x^{2} - 4 y^{2} - 8 = 0

What is the foci ((x+2)^2)/9-((y-1)^2)/(169)=1 ?
The foci ((x+2)^2)/9-((y-1)^2)/(169)=1 is (-2+sqrt(178),1),(-2-sqrt(178),1)