What is the eccentricity (x^2)/4-(y^2)/(16)=1 ?

The eccentricity (x^2)/4-(y^2)/(16)=1 is square root of 5

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eccentricity (x^2)/4-(y^2)/(16)=1

eccentricity

\frac{x ^{2}}{4} - \frac{y ^{2}}{16} = 1

5

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{4} - \frac{y ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 4

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

Refine

= 5

49 x^{2} - 25 y^{2} - 686 x - 50 y + 2375 = 0

\frac{y ^{2}}{25} - \frac{( x + 1 ) ^{2}}{16} = 1

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

\frac{y ^{2}}{36} - \frac{x ^{2}}{9} = 1

16 x^{2} - 25 y^{2} = 100

What is the eccentricity (x^2)/4-(y^2)/(16)=1 ?
The eccentricity (x^2)/4-(y^2)/(16)=1 is square root of 5