What is the foci (x^2)/4-((y-6)^2)/(140)=1 ?

The foci (x^2)/4-((y-6)^2)/(140)=1 is (12,6),(-12,6)

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foci (x^2)/4-((y-6)^2)/(140)=1

foci

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

(12, 6), (- 12, 6)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 6), a = 2, b = 235

(0 + c, 6), (0 - c, 6)

Compute

c :

c = 2^{2} + (235)^{2} : 12

(0 + 12, 6), (0 - 12, 6)

Refine

(12, 6), (- 12, 6)

\frac{y ^{2}}{5} - \frac{x ^{2}}{5} = 1

(x - 3)^{2} - \frac{y ^{2}}{9} = 1

\frac{1}{144} x^{2} - \frac{1}{196} y^{2} = 1

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

\frac{x ^{2}}{2} - \frac{y ^{2}}{3} = 1

What is the foci (x^2)/4-((y-6)^2)/(140)=1 ?
The foci (x^2)/4-((y-6)^2)/(140)=1 is (12,6),(-12,6)