What is the foci (x^2)/(60)-(y^2)/(40)=1 ?

The foci (x^2)/(60)-(y^2)/(40)=1 is (10,0),(-10,0)

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foci (x^2)/(60)-(y^2)/(40)=1

foci

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

(10, 0), (- 10, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 215, b = 210

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

Compute

c :

c = (215)^{2} + (210)^{2} : 10

(0 + 10, 0), (0 - 10, 0)

Refine

(10, 0), (- 10, 0)

\frac{( y + 2 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{64} = 1

\frac{x ^{2}}{37} - \frac{y ^{2}}{63} = 1

3 x^{2} - y^{2} = 300

4 x^{2} - y^{2} = 4

(x - 1)^{2} - (y - 2)^{2} = 3

What is the foci (x^2)/(60)-(y^2)/(40)=1 ?
The foci (x^2)/(60)-(y^2)/(40)=1 is (10,0),(-10,0)