What is the foci ((x-13)^2)/(144)-(y^2)/(25)=1 ?

The foci ((x-13)^2)/(144)-(y^2)/(25)=1 is (26,0),(0,0)

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foci ((x-13)^2)/(144)-(y^2)/(25)=1

foci

\frac{( x - 13 ) ^{2}}{144} - \frac{y ^{2}}{25} = 1

(26, 0), (0, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 13 ) ^{2}}{144} - \frac{y ^{2}}{25} = 1 :

Right-left Hyperbola with

(h, k) = (13, 0), a = 12, b = 5

(13 + c, 0), (13 - c, 0)

Compute

c :

c = 1 2^{2} + 5^{2} : 13

(13 + 13, 0), (13 - 13, 0)

Refine

(26, 0), (0, 0)

9 x^{2} - 4 y^{2} - 90 x - 32 y = - 305

\frac{( y - 2 ) ^{2}}{40} - \frac{( x - 4 ) ^{2}}{9} = 1

5 y^{2} - 16 x^{2} + 400 = 0

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1

\frac{( y - 1 ) ^{2}}{49} - \frac{( x - 2 ) ^{2}}{32} = 1

What is the foci ((x-13)^2)/(144)-(y^2)/(25)=1 ?
The foci ((x-13)^2)/(144)-(y^2)/(25)=1 is (26,0),(0,0)