What is the foci ((x+4)^2)/9-((y+3)^2)/(16)=1 ?

The foci ((x+4)^2)/9-((y+3)^2)/(16)=1 is (1,-3),(-9,-3)

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foci ((x+4)^2)/9-((y+3)^2)/(16)=1

foci

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

(1, - 3), (- 9, - 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1 :

Right-left Hyperbola with

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

(- 4 + c, - 3), (- 4 - c, - 3)

Compute

c :

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

(- 4 + 5, - 3), (- 4 - 5, - 3)

Refine

(1, - 3), (- 9, - 3)

- 4 y^{2} + 9 x^{2} - 27 x - 16 y = 2

\frac{x ^{2}}{4 ^{2}} - \frac{y ^{2}}{3 ^{2}} = 1

- 9 x^{2} + 4 y^{2} = 36

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

- \frac{x ^{2}}{16} + \frac{y ^{2}}{36} = 1

What is the foci ((x+4)^2)/9-((y+3)^2)/(16)=1 ?
The foci ((x+4)^2)/9-((y+3)^2)/(16)=1 is (1,-3),(-9,-3)