What is the foci (x^2)/4-(y^2)/5 =1 ?

The foci (x^2)/4-(y^2)/5 =1 is (3,0),(-3,0)

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foci (x^2)/4-(y^2)/5 =1

foci

\frac{x ^{2}}{4} - \frac{y ^{2}}{5} = 1

(3, 0), (- 3, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{4} - \frac{y ^{2}}{5} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 5

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

Compute

c :

c = 2^{2} + 5^{2} : 3

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

4 x^{2} - y^{2} - 24 x - 4 y + 28 = 0

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

7 x^{2} - 9 y^{2} - 14 x + 72 y - 200 = 0

x^{2} - 4 y^{2} = 4

\frac{x ^{2}}{25} - \frac{y ^{2}}{49} = 1

What is the foci (x^2)/4-(y^2)/5 =1 ?
The foci (x^2)/4-(y^2)/5 =1 is (3,0),(-3,0)