What is the foci ((x-1)^2}{25}-\frac{(y+3)^2)/9 =1 ?

The foci ((x-1)^2}{25}-\frac{(y+3)^2)/9 =1 is (1+sqrt(34),-3),(1-sqrt(34),-3)

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foci ((x-1)^2}{25}-\frac{(y+3)^2)/9 =1

foci

\frac{( x - 1 ) ^{2}}{25} - \frac{( y + 3 ) ^{2}}{9} = 1

(1 + 34, - 3), (1 - 34, - 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 1 ) ^{2}}{25} - \frac{( y + 3 ) ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (1, - 3), a = 5, b = 3

(1 + c, - 3), (1 - c, - 3)

Compute

c :

c = 5^{2} + 3^{2} : 34

(1 + 34, - 3), (1 - 34, - 3)

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

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

What is the foci ((x-1)^2}{25}-\frac{(y+3)^2)/9 =1 ?
The foci ((x-1)^2}{25}-\frac{(y+3)^2)/9 =1 is (1+sqrt(34),-3),(1-sqrt(34),-3)