What is the eccentricity 18x^2-32y^2-108y-128x-254=0 ?

The eccentricity 18x^2-32y^2-108y-128x-254=0 is 5/4

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eccentricity 18x^2-32y^2-108y-128x-254=0

eccentricity

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

\frac{5}{4}

Solution steps

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

Calculate Hyperbola properties

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0 :

Right-left Hyperbola with

(h, k) = (\frac{32}{9}, - \frac{27}{16}), a = \frac{28111}{36}, b = \frac{28111}{48}

= \frac{( \frac{28111}{36} ) ^{2} + ( \frac{28111}{48} ) ^{2}}{\frac{28111}{36}}

Refine

= \frac{5}{4}

25 x^{2} + 246 x - 25 y^{2} - 13 y = 233

9 x^{2} - y^{2} - 36 x - 6 y + 18 = 0

- 9 x^{2} + 8 y^{2} + 36 x + 48 y = 36

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

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

What is the eccentricity 18x^2-32y^2-108y-128x-254=0 ?
The eccentricity 18x^2-32y^2-108y-128x-254=0 is 5/4