What is the foci ((x-3)^2)/3-((y-2)^2)/(22)=1 ?

The foci ((x-3)^2)/3-((y-2)^2)/(22)=1 is (8,2),(-2,2)

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foci ((x-3)^2)/3-((y-2)^2)/(22)=1

foci

\frac{( x - 3 ) ^{2}}{3} - \frac{( y - 2 ) ^{2}}{22} = 1

(8, 2), (- 2, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{3} - \frac{( y - 2 ) ^{2}}{22} = 1 :

Right-left Hyperbola with

(h, k) = (3, 2), a = 3, b = 22

(3 + c, 2), (3 - c, 2)

Compute

c :

c = 3^{2} + 22^{2} : 5

(3 + 5, 2), (3 - 5, 2)

Refine

(8, 2), (- 2, 2)

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

9 y^{2} - 4 x^{2} - 36 y - 8 x = 4

x^{2} - 4 y^{2} = 36

\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1

y^{2} = x^{2} - 8

What is the foci ((x-3)^2)/3-((y-2)^2)/(22)=1 ?
The foci ((x-3)^2)/3-((y-2)^2)/(22)=1 is (8,2),(-2,2)