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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

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

Refine

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

\frac{y ^{2}}{25} - \frac{x ^{2}}{81} = 1

\frac{( x - 5 ) ^{2}}{16} - \frac{( y + 7 ) ^{2}}{9} = - 1

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

\frac{y ^{2}}{15} - \frac{x ^{2}}{1} = 1

x^{2} - y^{2} + 7 x + 8 y = 5

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