What is the foci ((x-7)^2)/(25)-(y^2)/(16)=1 ?

The foci ((x-7)^2)/(25)-(y^2)/(16)=1 is (7+sqrt(41),0),(7-sqrt(41),0)

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foci ((x-7)^2)/(25)-(y^2)/(16)=1

foci

\frac{( x - 7 ) ^{2}}{25} - \frac{y ^{2}}{16} = 1

(7 + 41, 0), (7 - 41, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 7 ) ^{2}}{25} - \frac{y ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (7, 0), a = 5, b = 4

(7 + c, 0), (7 - c, 0)

Compute

c :

c = 5^{2} + 4^{2} : 41

(7 + 41, 0), (7 - 41, 0)

\frac{x ^{2}}{5} - \frac{y ^{2}}{7} = 1

\frac{( x - 6 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{9} = 1

3 x^{2} - 2 y^{2} = 1

\frac{y ^{2}}{25} - \frac{x ^{2}}{25} = 1

\frac{( x - 6 ) ^{2}}{10} - \frac{( y + 2 ) ^{2}}{17} = 1

What is the foci ((x-7)^2)/(25)-(y^2)/(16)=1 ?
The foci ((x-7)^2)/(25)-(y^2)/(16)=1 is (7+sqrt(41),0),(7-sqrt(41),0)