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

The foci ((x-2)^2)/9-((y-1)^2)/(16)=1 is (7,1),(-3,1)

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

foci

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

(7, 1), (- 3, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

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

Refine

(7, 1), (- 3, 1)

\frac{( x + 1 ) ^{2}}{25} - \frac{( y - 4 ) ^{2}}{144} = 1

y^{2} - x^{2} = 36

y^{2} - x^{2} = 22

16 x^{2} - 96 x - 9 y^{2} + 36 y - 36 = 0

\frac{x ^{2}}{4} - \frac{y ^{2}}{49} = 1

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