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

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

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

foci

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

(6, 2), (- 4, 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (1, 2), a = 3, b = 4

(1 + c, 2), (1 - c, 2)

Compute

c :

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

(1 + 5, 2), (1 - 5, 2)

Refine

(6, 2), (- 4, 2)

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

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

\frac{x ^{2}}{100} - \frac{y ^{2}}{49} = 1

\frac{( y - ( - 2 ) ) ^{2}}{9} - \frac{( x - 5 ) ^{2}}{27} = 1

\frac{( y - 0 ) ^{2}}{1 ^{2}} - \frac{( 3 x - 0 ) ^{2}}{90} = 1

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