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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{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 ^{2}}{34} - \frac{y ^{2}}{2} = 1

16 y^{2} - 49 x^{2} = 784

\frac{( y + 2 ) ^{2}}{25} - \frac{( x - 9 ) ^{2}}{81} = 1

4 x^{2} - 9 y^{2} + 32 x - 18 y + 37 = 0

- 4 x^{2} + y^{2} + 40 x - 136 = 0

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