What is the foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1 ?

The foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1 is (11,2),(1,2)

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foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1

foci

\frac{( x - 6 ) ^{2}}{16} - \frac{( y - 2 ) ^{2}}{9} = 1

(11, 2), (1, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 6 ) ^{2}}{16} - \frac{( y - 2 ) ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (6, 2), a = 4, b = 3

(6 + c, 2), (6 - c, 2)

Compute

c :

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

(6 + 5, 2), (6 - 5, 2)

Refine

(11, 2), (1, 2)

\frac{y ^{2}}{16} - \frac{x ^{2}}{25} = 1

- x^{2} + 4 y^{2} - 2 x - 16 y + 11 = 0

9 x^{2} - 16 y^{2} - 18 x - 64 y - 199 = 0

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

3 x^{2} - 15 y^{2} + 2 x - 3 y = 4

What is the foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1 ?
The foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1 is (11,2),(1,2)