What is the foci ((x+1)^2)/(25)-((y-4)^2)/(144)=1 ?

The foci ((x+1)^2)/(25)-((y-4)^2)/(144)=1 is (12,4),(-14,4)

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foci ((x+1)^2)/(25)-((y-4)^2)/(144)=1

foci

\frac{( x + 1 ) ^{2}}{25} - \frac{( y - 4 ) ^{2}}{144} = 1

(12, 4), (- 14, 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 1 ) ^{2}}{25} - \frac{( y - 4 ) ^{2}}{144} = 1 :

Right-left Hyperbola with

(h, k) = (- 1, 4), a = 5, b = 12

(- 1 + c, 4), (- 1 - c, 4)

Compute

c :

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

(- 1 + 13, 4), (- 1 - 13, 4)

Refine

(12, 4), (- 14, 4)

y^{2} - x^{2} = 36

y^{2} - x^{2} = 22

16 x^{2} - 96 x - 9 y^{2} + 36 y - 36 = 0

\frac{x ^{2}}{4} - \frac{y ^{2}}{49} = 1

\frac{( y - 1 ) ^{2}}{4} - \frac{( x + 8 ) ^{2}}{25} = 1

What is the foci ((x+1)^2)/(25)-((y-4)^2)/(144)=1 ?
The foci ((x+1)^2)/(25)-((y-4)^2)/(144)=1 is (12,4),(-14,4)