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

The eccentricity (x^2}{16}-\frac{y^2)/9 =1 is 5/4

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eccentricity (x^2}{16}-\frac{y^2)/9 =1

eccentricity

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

\frac{5}{4}

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

= \frac{5}{4}

\frac{y ^{2}}{25} - \frac{x ^{2}}{81} = 1

- x^{2} + 9 y^{2} + 4 x + 54 y + 68 = 0

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

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

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

What is the eccentricity (x^2}{16}-\frac{y^2)/9 =1 ?
The eccentricity (x^2}{16}-\frac{y^2)/9 =1 is 5/4