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

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

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

foci

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

(2 + 10, - 3), (2 - 10, - 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

(2 + 10, - 3), (2 - 10, - 3)

\frac{( x + 5 ) ^{2}}{36} - \frac{( y + 1 ) ^{2}}{64} = 1

x^{2} - y^{2} - 4 x - 2 y - 1 = 0

9 x^{2} - y^{2} - 54 x + 45 = 0

x^{2} - 16 y^{2} = 16

3 x^{2} - y^{2} = 11

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