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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

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

Refine

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

\frac{( x + 2 ) ^{2}}{49} - \frac{( y - 2 ) ^{2}}{45} = 1

\frac{( y - 3 ) ^{2}}{8 ^{2}} - \frac{( x - 2 ) ^{2}}{8 ^{2}} = 1

\frac{x ^{2}}{18} - \frac{y ^{2}}{7} = 1

\frac{( x ^{2} )}{4} - 0.14 y^{2} = 1

\frac{x ^{2}}{( \frac{64}{7} )} - \frac{y ^{2}}{( \frac{48}{7} )} = 1

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