What is the foci ((x-3)^2}{13}-\frac{(y-5)^2)/7 =1 ?

The foci ((x-3)^2}{13}-\frac{(y-5)^2)/7 =1 is (3+2sqrt(5),5),(3-2sqrt(5),5)

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foci ((x-3)^2}{13}-\frac{(y-5)^2)/7 =1

foci

\frac{( x - 3 ) ^{2}}{13} - \frac{( y - 5 ) ^{2}}{7} = 1

(3 + 25, 5), (3 - 25, 5)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{13} - \frac{( y - 5 ) ^{2}}{7} = 1 :

Right-left Hyperbola with

(h, k) = (3, 5), a = 13, b = 7

(3 + c, 5), (3 - c, 5)

Compute

c :

c = 13^{2} + 7^{2} : 25

(3 + 25, 5), (3 - 25, 5)

\frac{y ^{2}}{10} - \frac{( x - 1 ) ^{2}}{10} = 1

- 4 x^{2} + y^{2} + 6 y - 27 = 0

4 y^{2} - 9 x^{2} + 8 y - 54 x - 113 = 0

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

\frac{( x - 7 ) ^{2}}{100} - \frac{( y + 8 ) ^{2}}{25} = 1

What is the foci ((x-3)^2}{13}-\frac{(y-5)^2)/7 =1 ?
The foci ((x-3)^2}{13}-\frac{(y-5)^2)/7 =1 is (3+2sqrt(5),5),(3-2sqrt(5),5)