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

The foci (x-1)^2-(y-2)^2=3 is (1+sqrt(6),2),(1-sqrt(6),2)

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

foci

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

(1 + 6, 2), (1 - 6, 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

c = 3^{2} + 3^{2} : 6

(1 + 6, 2), (1 - 6, 2)

4 x^{2} - y^{2} - 6 y - 25 = 0

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

16 x^{2} - y^{2} - 64 x - 2 y + 47 = 0

\frac{( x - 1 ) ^{2}}{16} - \frac{( y - 7 ) ^{2}}{20} = 1

y^{2} - 9 x^{2} - 8 y - 54 x - 81 = 0

What is the foci (x-1)^2-(y-2)^2=3 ?
The foci (x-1)^2-(y-2)^2=3 is (1+sqrt(6),2),(1-sqrt(6),2)