What is the foci ((x+2)^2)/9-((y-1)^2)/(16)=1 ?

The foci ((x+2)^2)/9-((y-1)^2)/(16)=1 is (3,1),(-7,1)

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foci ((x+2)^2)/9-((y-1)^2)/(16)=1

foci

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

(3, 1), (- 7, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

c = 3^{2} + 4^{2} : 5

(- 2 + 5, 1), (- 2 - 5, 1)

Refine

(3, 1), (- 7, 1)

\frac{( x + 4 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{25} = 1

\frac{( y + 4 ) ^{2}}{9} - \frac{( x - 3 ) ^{2}}{25} = 1

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

- \frac{( y - 3 ) ^{2}}{25} + \frac{( x + 2 ) ^{2}}{9} = 1

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 5 ) ^{2}}{9} = 1

What is the foci ((x+2)^2)/9-((y-1)^2)/(16)=1 ?
The foci ((x+2)^2)/9-((y-1)^2)/(16)=1 is (3,1),(-7,1)