What is the foci ((x-1)^2}{12}-\frac{(y-2)^2)/8 =1 ?

The foci ((x-1)^2}{12}-\frac{(y-2)^2)/8 =1 is (1+2sqrt(5),2),(1-2sqrt(5),2)

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foci ((x-1)^2}{12}-\frac{(y-2)^2)/8 =1

foci

\frac{( x - 1 ) ^{2}}{12} - \frac{( y - 2 ) ^{2}}{8} = 1

(1 + 25, 2), (1 - 25, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 1 ) ^{2}}{12} - \frac{( y - 2 ) ^{2}}{8} = 1 :

Right-left Hyperbola with

(h, k) = (1, 2), a = 23, b = 22

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

Compute

c :

c = (23)^{2} + (22)^{2} : 25

(1 + 25, 2), (1 - 25, 2)

- x^{2} + 25 \cdot y^{2} = 25

4 x^{2} - y^{2} + 12 x + 5 = 0

\frac{( x - 13 ) ^{2}}{144} - \frac{y ^{2}}{25} = 1

9 x^{2} - 4 y^{2} - 90 x - 32 y = - 305

\frac{( y - 2 ) ^{2}}{40} - \frac{( x - 4 ) ^{2}}{9} = 1

What is the foci ((x-1)^2}{12}-\frac{(y-2)^2)/8 =1 ?
The foci ((x-1)^2}{12}-\frac{(y-2)^2)/8 =1 is (1+2sqrt(5),2),(1-2sqrt(5),2)