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

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

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

eccentricity

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

\frac{14}{5}

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

= \frac{14}{5}

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

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

\frac{x ^{2}}{45} - \frac{y ^{2}}{216} = 1

16 x^{2} - 9 y^{2} - 64 x - 18 y + 199 = 0

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

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