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

The foci ((x+4)^2)/4-((y-1)^2)/9 =1 is (-4+sqrt(13),1),(-4-sqrt(13),1)

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

foci

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

(- 4 + 13, 1), (- 4 - 13, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 4, 1), a = 2, b = 3

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

Compute

c :

c = 2^{2} + 3^{2} : 13

(- 4 + 13, 1), (- 4 - 13, 1)

\frac{x ^{2}}{16} - \frac{y ^{2}}{49} = 1

x^{2} - 6 x - 6 = y^{2} + 6 y + 6

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

- 4 y^{2} + 9 x^{2} - 27 x - 16 y = 2

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

What is the foci ((x+4)^2)/4-((y-1)^2)/9 =1 ?
The foci ((x+4)^2)/4-((y-1)^2)/9 =1 is (-4+sqrt(13),1),(-4-sqrt(13),1)