What is the foci ((x^2))/4-0.2y^2=1 ?

The foci ((x^2))/4-0.2y^2=1 is (3,0),(-3,0)

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foci ((x^2))/4-0.2y^2=1

foci

\frac{( x ^{2} )}{4} - 0.2 y^{2} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{4} - 0.2 y^{2} = 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)

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 4 ) ^{2}}{49} = 1

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

\frac{( y + 2 ) ^{2}}{4} - \frac{( x - 1 ) ^{2}}{16} = 1

x^{2} - 2 x - 3 = y^{2} + 2 y + 3

6 x^{2} - 5 y^{2} - 24 x + 10 y - 11 = 0

What is the foci ((x^2))/4-0.2y^2=1 ?
The foci ((x^2))/4-0.2y^2=1 is (3,0),(-3,0)