What is the eccentricity (x^2)/(16)-(y^2)/(20)=1 ?

The eccentricity (x^2)/(16)-(y^2)/(20)=1 is 3/2

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eccentricity (x^2)/(16)-(y^2)/(20)=1

eccentricity

\frac{x ^{2}}{16} - \frac{y ^{2}}{20} = 1

\frac{3}{2}

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{20} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 25

= \frac{4 ^{2} + ( 2 5 ) ^{2}}{4}

Refine

= \frac{3}{2}

16 x^{2} - 25 y^{2} = 400

16 x^{2} - 9 y^{2} = 144

4 x^{2} - y^{2} - 24 x - 4 y + 28 = 0

9 y^{2} - x^{2} + 2 x + 54 y + 62 = 0

4 x^{2} - 25 y^{2} + 100 y = 200

What is the eccentricity (x^2)/(16)-(y^2)/(20)=1 ?
The eccentricity (x^2)/(16)-(y^2)/(20)=1 is 3/2