What is the foci (x^2)/(64)-(y^2)/(32)=1 ?

The foci (x^2)/(64)-(y^2)/(32)=1 is (4sqrt(6),0),(-4sqrt(6),0)

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foci (x^2)/(64)-(y^2)/(32)=1

foci

\frac{x ^{2}}{64} - \frac{y ^{2}}{32} = 1

(46, 0), (- 46, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{64} - \frac{y ^{2}}{32} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 8, b = 42

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

Compute

c :

c = 8^{2} + (42)^{2} : 46

(0 + 46, 0), (0 - 46, 0)

Refine

(46, 0), (- 46, 0)

25 y^{2} - 81 x^{2} = 2025

\frac{y ^{2}}{4} - \frac{x ^{2}}{16} = 1

\frac{( y + 3 ) ^{2}}{64} - \frac{( x + 6 ) ^{2}}{49} = 1

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

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

What is the foci (x^2)/(64)-(y^2)/(32)=1 ?
The foci (x^2)/(64)-(y^2)/(32)=1 is (4sqrt(6),0),(-4sqrt(6),0)