What is the eccentricity (x^2)/9-(y^2)/(16)=1 ?

The eccentricity (x^2)/9-(y^2)/(16)=1 is 5/3

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eccentricity (x^2)/9-(y^2)/(16)=1

eccentricity

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

\frac{5}{3}

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 4

= \frac{3 ^{2} + 4 ^{2}}{3}

Refine

= \frac{5}{3}

- 9 x^{2} + y^{2} - 72 x - 153 = 0

4 x^{2} - y^{2} = 16

16 x^{2} - 96 x - y^{2} - 2 y + 139 = 0

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

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

What is the eccentricity (x^2)/9-(y^2)/(16)=1 ?
The eccentricity (x^2)/9-(y^2)/(16)=1 is 5/3