What is the eccentricity y^2-4y-4x^2+13=0 ?

The eccentricity y^2-4y-4x^2+13=0 is square root of 5

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eccentricity y^2-4y-4x^2+13=0

eccentricity

y^{2} - 4 y - 4 x^{2} + 13 = 0

5

Solution steps

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

Calculate Hyperbola properties

y^{2} - 4 y - 4 x^{2} + 13 = 0 :

Right-left Hyperbola with

(h, k) = (0, 2), a = \frac{3}{2}, b = 3

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

Refine

= 5

+ \frac{y ^{2}}{25} - \frac{x ^{2}}{4} = 1

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

25 x^{2} + 56476 x - 25 y^{2} - 3 y = 48838

\frac{x ^{2}}{25} - \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{40} - \frac{y ^{2}}{174} = 1

What is the eccentricity y^2-4y-4x^2+13=0 ?
The eccentricity y^2-4y-4x^2+13=0 is square root of 5