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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

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

Refine

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

x^{2} - 6 x - 6 = y^{2} + 6 y + 6

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

- 4 y^{2} + 9 x^{2} - 27 x - 16 y = 2

\frac{x ^{2}}{4 ^{2}} - \frac{y ^{2}}{3 ^{2}} = 1

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

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