What is the foci (x^2}{36}-\frac{y^2)/4 =1 ?

The foci (x^2}{36}-\frac{y^2)/4 =1 is (2sqrt(10),0),(-2sqrt(10),0)

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foci (x^2}{36}-\frac{y^2)/4 =1

foci

\frac{x ^{2}}{36} - \frac{y ^{2}}{4} = 1

(210, 0), (- 210, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{36} - \frac{y ^{2}}{4} = 1 :

Right-left Hyperbola with

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

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

Compute

c :

c = 6^{2} + 2^{2} : 210

(0 + 210, 0), (0 - 210, 0)

Refine

(210, 0), (- 210, 0)

- 9 x^{2} - 32 y = - 16 y^{2} + 128

\frac{x ^{2}}{16} - \frac{y ^{2}}{16} = 1

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{16} = 1

\frac{x ^{2}}{25} - \frac{y ^{2}}{64} = 1

What is the foci (x^2}{36}-\frac{y^2)/4 =1 ?
The foci (x^2}{36}-\frac{y^2)/4 =1 is (2sqrt(10),0),(-2sqrt(10),0)