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

The eccentricity (x^2)/(60)-(y^2)/(40)=1 is sqrt(5/3)

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

eccentricity

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

\frac{5}{3}

Solution steps

= \frac{a ^{2} + b ^{2}}{a}

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

= \frac{( 2 15 ) ^{2} + ( 2 10 ) ^{2}}{2 15}

Refine

= \frac{5}{3}

x^{2} - 3 y^{2} - 2 x - 6 y + 7 = 0

\frac{x ^{2}}{16} - \frac{y ^{2}}{64} = 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{12} = 1

x^{2} - y^{2} < 1

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

What is the eccentricity (x^2)/(60)-(y^2)/(40)=1 ?
The eccentricity (x^2)/(60)-(y^2)/(40)=1 is sqrt(5/3)