What is the foci ((x-0)^2}{16}-\frac{(y-0)^2)/9 =1 ?

The foci ((x-0)^2}{16}-\frac{(y-0)^2)/9 =1 is (5,0),(-5,0)

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foci ((x-0)^2}{16}-\frac{(y-0)^2)/9 =1

foci

\frac{( x - 0 ) ^{2}}{16} - \frac{( y - 0 ) ^{2}}{9} = 1

(5, 0), (- 5, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 0 ) ^{2}}{16} - \frac{( y - 0 ) ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 3

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

Compute

c :

c = 4^{2} + 3^{2} : 5

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

\frac{( x - 4 ) ^{2}}{49} - \frac{( y + 6 ) ^{2}}{9} = 1

\frac{( x ^{2} )}{36} - \frac{( y ^{2} )}{25} = 1

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

y^{2} - 4 x^{2} < 1

x^{2} - 16 y^{2} = 64

What is the foci ((x-0)^2}{16}-\frac{(y-0)^2)/9 =1 ?
The foci ((x-0)^2}{16}-\frac{(y-0)^2)/9 =1 is (5,0),(-5,0)