What is the foci ((x-3)^2)/(81)-((y+5)^2)/(144)=1 ?

The foci ((x-3)^2)/(81)-((y+5)^2)/(144)=1 is (18,-5),(-12,-5)

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foci ((x-3)^2)/(81)-((y+5)^2)/(144)=1

foci

\frac{( x - 3 ) ^{2}}{81} - \frac{( y + 5 ) ^{2}}{144} = 1

(18, - 5), (- 12, - 5)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{81} - \frac{( y + 5 ) ^{2}}{144} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 5), a = 9, b = 12

(3 + c, - 5), (3 - c, - 5)

Compute

c :

c = 9^{2} + 1 2^{2} : 15

(3 + 15, - 5), (3 - 15, - 5)

Refine

(18, - 5), (- 12, - 5)

\frac{x ^{2}}{1} - \frac{y ^{2}}{9} = 1

y^{2} - x^{2} = 25

\frac{x ^{2}}{140} - \frac{( y - 6 ) ^{2}}{4} = - 1

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

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

What is the foci ((x-3)^2)/(81)-((y+5)^2)/(144)=1 ?
The foci ((x-3)^2)/(81)-((y+5)^2)/(144)=1 is (18,-5),(-12,-5)