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

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

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

foci

(25, - 1), (- 25, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, - 1), a = 2, b = 4

(0 + c, - 1), (0 - c, - 1)

Compute

c :

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

(0 + 25, - 1), (0 - 25, - 1)

Refine

(25, - 1), (- 25, - 1)

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