What is the foci (x^2)/(25)-(y^2)/(49)=1 ?

The foci (x^2)/(25)-(y^2)/(49)=1 is (sqrt(74),0),(-sqrt(74),0)

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foci (x^2)/(25)-(y^2)/(49)=1

foci

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

(74, 0), (- 74, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 7

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

Compute

c :

c = 5^{2} + 7^{2} : 74

(0 + 74, 0), (0 - 74, 0)

Refine

(74, 0), (- 74, 0)

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

\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

What is the foci (x^2)/(25)-(y^2)/(49)=1 ?
The foci (x^2)/(25)-(y^2)/(49)=1 is (sqrt(74),0),(-sqrt(74),0)