What is the foci (x^2)/4-(y^2)/(25)=1 ?

The foci (x^2)/4-(y^2)/(25)=1 is (sqrt(29),0),(-sqrt(29),0)

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foci (x^2)/4-(y^2)/(25)=1

foci

\frac{x ^{2}}{4} - \frac{y ^{2}}{25} = 1

(29, 0), (- 29, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{4} - \frac{y ^{2}}{25} = 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} : 29

(0 + 29, 0), (0 - 29, 0)

Refine

(29, 0), (- 29, 0)

\frac{y ^{2}}{1} - \frac{x ^{2}}{15} = 1

\frac{( x + 4 ) ^{2}}{4} - \frac{( y - 1 ) ^{2}}{25} = 1

9 (x - 1)^{2} - 16 (y + 2)^{2} = 144

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

\frac{x ^{2}}{9} - \frac{y ^{2}}{40} = 1

What is the foci (x^2)/4-(y^2)/(25)=1 ?
The foci (x^2)/4-(y^2)/(25)=1 is (sqrt(29),0),(-sqrt(29),0)