What is the foci (x^2)/(16)-(y^2)/(48)=1 ?

The foci (x^2)/(16)-(y^2)/(48)=1 is (8,0),(-8,0)

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foci (x^2)/(16)-(y^2)/(48)=1

foci

\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1

(8, 0), (- 8, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 43

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

Compute

c :

c = 4^{2} + (43)^{2} : 8

(0 + 8, 0), (0 - 8, 0)

Refine

(8, 0), (- 8, 0)

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

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{18} = 1

\frac{( y - 3 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{27} = 1

y^{2} - 4 y - 4 x^{2} + 13 = 0

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 10 ) ^{2}}{100} = 1

What is the foci (x^2)/(16)-(y^2)/(48)=1 ?
The foci (x^2)/(16)-(y^2)/(48)=1 is (8,0),(-8,0)