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

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

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

foci

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

(1 + 41, 2), (1 - 41, 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (1, 2), a = 5, b = 4

(1 + c, 2), (1 - c, 2)

Compute

c :

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

(1 + 41, 2), (1 - 41, 2)

9 x^{2} - 16 y^{2} - 108 x + 64 y + 116 = 0

y^{2} - x^{2} = \frac{9}{25}

\frac{y ^{2}}{25} - \frac{x ^{2}}{11} = 1

\frac{( y ) ^{2}}{121} - \frac{( x ) ^{2}}{81} = 1

8 x^{2} - 3 y^{2} = 24

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