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

The foci (x^2)/(16)-(y^2)/(16)=1 is (4sqrt(2),0),(-4sqrt(2),0)

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

foci

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

(42, 0), (- 42, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

c = 4^{2} + 4^{2} : 42

(0 + 42, 0), (0 - 42, 0)

Refine

(42, 0), (- 42, 0)

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

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

\frac{x ^{2}}{25} - \frac{y ^{2}}{64} = 1

\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1

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

What is the foci (x^2)/(16)-(y^2)/(16)=1 ?
The foci (x^2)/(16)-(y^2)/(16)=1 is (4sqrt(2),0),(-4sqrt(2),0)