What is the foci ((x+3)^2)/4-((y+1)^2)/(16)=1 ?

The foci ((x+3)^2)/4-((y+1)^2)/(16)=1 is (-3+2sqrt(5),-1),(-3-2sqrt(5),-1)

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foci ((x+3)^2)/4-((y+1)^2)/(16)=1

foci

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

(- 3 + 25, - 1), (- 3 - 25, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(- 3 + c, - 1), (- 3 - c, - 1)

Compute

c :

c = 2^{2} + 4^{2} : 25

(- 3 + 25, - 1), (- 3 - 25, - 1)

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

- 4 x^{2} + y^{2} - 16 x - 52 = 0

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

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 4 ) ^{2}}{25} = 1

4 x^{2} - 8 y^{2} = 3

What is the foci ((x+3)^2)/4-((y+1)^2)/(16)=1 ?
The foci ((x+3)^2)/4-((y+1)^2)/(16)=1 is (-3+2sqrt(5),-1),(-3-2sqrt(5),-1)