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

The foci ((x-1)^2)/9-((y-3)^2)/(16)=1 is (6,3),(-4,3)

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

foci

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

(6, 3), (- 4, 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(1 + c, 3), (1 - c, 3)

Compute

c :

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

(1 + 5, 3), (1 - 5, 3)

Refine

(6, 3), (- 4, 3)

2 x^{2} - 7 y^{2} - 16 x + 14 y = - 11

9 x^{2} - 4 y^{2} - 36 x + 8 y - 4 = 0

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

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

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

What is the foci ((x-1)^2)/9-((y-3)^2)/(16)=1 ?
The foci ((x-1)^2)/9-((y-3)^2)/(16)=1 is (6,3),(-4,3)