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

The foci ((x+3)^2)/9-((y-2)^2)/(16)=1 is (2,2),(-8,2)

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foci ((x+3)^2)/9-((y-2)^2)/(16)=1

foci

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

(2, 2), (- 8, 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(- 3 + c, 2), (- 3 - c, 2)

Compute

c :

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

(- 3 + 5, 2), (- 3 - 5, 2)

Refine

(2, 2), (- 8, 2)

9 x^{2} - y^{2} - 54 x - 2 y - 1 = 0

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

\frac{( y + 2 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{64} = 1

\frac{x ^{2}}{37} - \frac{y ^{2}}{63} = 1

3 x^{2} - y^{2} = 300

What is the foci ((x+3)^2)/9-((y-2)^2)/(16)=1 ?
The foci ((x+3)^2)/9-((y-2)^2)/(16)=1 is (2,2),(-8,2)