What is the eccentricity (x^2)/(144)-(y^2)/(25)=1 ?

The eccentricity (x^2)/(144)-(y^2)/(25)=1 is 13/12

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eccentricity (x^2)/(144)-(y^2)/(25)=1

eccentricity

\frac{x ^{2}}{144} - \frac{y ^{2}}{25} = 1

\frac{13}{12}

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{144} - \frac{y ^{2}}{25} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 12, b = 5

= \frac{1 2 ^{2} + 5 ^{2}}{12}

Refine

= \frac{13}{12}

9 (x - 2)^{2} - 4 y^{2} = 36

- \frac{( x - 2 ) ^{2}}{5} + \frac{( y + 1 ) ^{2}}{9} = 1

\frac{x ^{2}}{58} - \frac{y ^{2}}{174} = 1

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

25 x^{2} + 10922 x - 25 y^{2} - y^{2} = 1439

What is the eccentricity (x^2)/(144)-(y^2)/(25)=1 ?
The eccentricity (x^2)/(144)-(y^2)/(25)=1 is 13/12