What is the foci ((x+4)^2)/4-((y-1)^2)/(25)=1 ?

The foci ((x+4)^2)/4-((y-1)^2)/(25)=1 is (-4+sqrt(29),1),(-4-sqrt(29),1)

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foci ((x+4)^2)/4-((y-1)^2)/(25)=1

foci

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

(- 4 + 29, 1), (- 4 - 29, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 4, 1), a = 2, b = 5

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

Compute

c :

c = 2^{2} + 5^{2} : 29

(- 4 + 29, 1), (- 4 - 29, 1)

9 (x - 1)^{2} - 16 (y + 2)^{2} = 144

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

\frac{x ^{2}}{9} - \frac{y ^{2}}{40} = 1

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

\frac{x ^{2}}{9} - \frac{y ^{2}}{1} = 1

What is the foci ((x+4)^2)/4-((y-1)^2)/(25)=1 ?
The foci ((x+4)^2)/4-((y-1)^2)/(25)=1 is (-4+sqrt(29),1),(-4-sqrt(29),1)