What is the foci ((x-6)/4)^2-(y/4)^2=1 ?

The foci ((x-6)/4)^2-(y/4)^2=1 is (6+4sqrt(2),0),(6-4sqrt(2),0)

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foci ((x-6)/4)^2-(y/4)^2=1

foci

(\frac{x - 6}{4})^{2} - (\frac{y}{4})^{2} = 1

(6 + 42, 0), (6 - 42, 0)

Solution steps

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

Calculate Hyperbola properties

(\frac{x - 6}{4})^{2} - (\frac{y}{4})^{2} = 1 :

Right-left Hyperbola with

(h, k) = (6, 0), a = 4, b = 4

(6 + c, 0), (6 - c, 0)

Compute

c :

c = 4^{2} + 4^{2} : 42

(6 + 42, 0), (6 - 42, 0)

25 y^{2} - 4 x^{2} = 100

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

4 x^{2} - 9 y^{2} + 3 x + 18 y + 91 = 0

4 x^{2} - 25 y^{2} = 100

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

What is the foci ((x-6)/4)^2-(y/4)^2=1 ?
The foci ((x-6)/4)^2-(y/4)^2=1 is (6+4sqrt(2),0),(6-4sqrt(2),0)