What is the foci ((x-3)^2)/(25)-((y+1)^2)/(25)=1 ?

The foci ((x-3)^2)/(25)-((y+1)^2)/(25)=1 is (3+5sqrt(2),-1),(3-5sqrt(2),-1)

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foci ((x-3)^2)/(25)-((y+1)^2)/(25)=1

foci

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

(3 + 52, - 1), (3 - 52, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

c = 5^{2} + 5^{2} : 52

(3 + 52, - 1), (3 - 52, - 1)

\frac{x ^{2}}{5625} - \frac{y ^{2}}{1600} = 1

\frac{( y - 6 ) ^{2}}{36} - \frac{( x + 1 ) ^{2}}{16} = 1

9 x^{2} - 4 y^{2} + 54 x + 16 y + 29 = 0

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

\frac{x ^{2}}{49} - \frac{y ^{2}}{4} = 1

What is the foci ((x-3)^2)/(25)-((y+1)^2)/(25)=1 ?
The foci ((x-3)^2)/(25)-((y+1)^2)/(25)=1 is (3+5sqrt(2),-1),(3-5sqrt(2),-1)