What is the eccentricity (x^2)/(40)-(y^2)/(120)=1 ?

The eccentricity (x^2)/(40)-(y^2)/(120)=1 is 2

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eccentricity (x^2)/(40)-(y^2)/(120)=1

eccentricity

\frac{x ^{2}}{40} - \frac{y ^{2}}{120} = 1

2

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{40} - \frac{y ^{2}}{120} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 210, b = 230

= \frac{( 2 10 ) ^{2} + ( 2 30 ) ^{2}}{2 10}

Refine

= 2

0 = - \frac{400}{9} x^{2} + 16 y^{2} + 400

\frac{x ^{2}}{121} - \frac{y ^{2}}{81} = 1

\frac{x ^{2}}{27} - \frac{y ^{2}}{216} = 1

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

9 x^{2} - 25 y^{2} = 225

What is the eccentricity (x^2)/(40)-(y^2)/(120)=1 ?
The eccentricity (x^2)/(40)-(y^2)/(120)=1 is 2