What is the eccentricity ((x-2)^2)/9-((y+1)^2)/5 =1 ?

The eccentricity ((x-2)^2)/9-((y+1)^2)/5 =1 is (sqrt(14))/3

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eccentricity ((x-2)^2)/9-((y+1)^2)/5 =1

eccentricity

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

\frac{14}{3}

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (2, - 1), a = 3, b = 5

= \frac{3 ^{2} + ( 5 ) ^{2}}{3}

Refine

= \frac{14}{3}

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

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

What is the eccentricity ((x-2)^2)/9-((y+1)^2)/5 =1 ?
The eccentricity ((x-2)^2)/9-((y+1)^2)/5 =1 is (sqrt(14))/3