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

The foci ((x-8)^2)/9-((y+3)^2)/(16)=1 is (13,-3),(3,-3)

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

foci

\frac{( x - 8 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

(13, - 3), (3, - 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 8 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (8, - 3), a = 3, b = 4

(8 + c, - 3), (8 - c, - 3)

Compute

c :

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

(8 + 5, - 3), (8 - 5, - 3)

Refine

(13, - 3), (3, - 3)

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

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

x^{2} - 9 y^{2} = 9

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

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

What is the foci ((x-8)^2)/9-((y+3)^2)/(16)=1 ?
The foci ((x-8)^2)/9-((y+3)^2)/(16)=1 is (13,-3),(3,-3)