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

The foci (x-3)^2-(y^2)/9 =1 is (3+sqrt(10),0),(3-sqrt(10),0)

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

foci

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

(3 + 10, 0), (3 - 10, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (3, 0), a = 1, b = 3

(3 + c, 0), (3 - c, 0)

Compute

c :

c = 1^{2} + 3^{2} : 10

(3 + 10, 0), (3 - 10, 0)

\frac{1}{144} x^{2} - \frac{1}{196} y^{2} = 1

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

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

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

9 x^{2} - 4 y^{2} - 18 x - 16 y - 43 = 0

What is the foci (x-3)^2-(y^2)/9 =1 ?
The foci (x-3)^2-(y^2)/9 =1 is (3+sqrt(10),0),(3-sqrt(10),0)