What is the foci (x^2)/(49)-(y^2)/(16)=1 ?

The foci (x^2)/(49)-(y^2)/(16)=1 is (sqrt(65),0),(-sqrt(65),0)

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foci (x^2)/(49)-(y^2)/(16)=1

foci

\frac{x ^{2}}{49} - \frac{y ^{2}}{16} = 1

(65, 0), (- 65, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{49} - \frac{y ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 7, b = 4

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

Compute

c :

c = 7^{2} + 4^{2} : 65

(0 + 65, 0), (0 - 65, 0)

Refine

(65, 0), (- 65, 0)

16 x^{2} + 25 y^{2} = 400

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

9 x^{2} - 4 y^{2} = 36

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

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

What is the foci (x^2)/(49)-(y^2)/(16)=1 ?
The foci (x^2)/(49)-(y^2)/(16)=1 is (sqrt(65),0),(-sqrt(65),0)