What is the foci 4x^2=y^2+4 ?

The foci 4x^2=y^2+4 is (sqrt(5),0),(-sqrt(5),0)

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foci 4x^2=y^2+4

foci

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

(5, 0), (- 5, 0)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} = y^{2} + 4 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 1, b = 2

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

Compute

c :

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

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

x^{2} - 4 y^{2} - 2 x + 16 y = 20

\frac{x ^{2}}{45} - \frac{y ^{2}}{360} = 1

\frac{y ^{2}}{225} - \frac{x ^{2}}{64} = 1

4 y^{2} - 9 x^{2} + 16 y + 18 x = 29

\frac{( x - 6 ) ^{2}}{4} - \frac{( y + 6 ) ^{2}}{81} = 1