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

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

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

eccentricity

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

5

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, - 1), a = 2, b = 4

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

Refine

= 5

x^{2} - 4 y^{2} - 2 x + 16 y = 20

5 y^{2} - 16 x^{2} + 400 = 0

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

\frac{x ^{2}}{25} - \frac{y ^{2}}{36} = 1

x^{2} - 4 y^{2} - 4 x + 8 y - 16 = 0

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