What is the eccentricity (x^2)/5+(y^2)/8 =1 ?

The eccentricity (x^2)/5+(y^2)/8 =1 is (sqrt(3))/(2sqrt(2))

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eccentricity (x^2)/5+(y^2)/8 =1

eccentricity

\frac{x ^{2}}{5} + \frac{y ^{2}}{8} = 1

\frac{3}{2 2}

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{5} + \frac{y ^{2}}{8} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 22, a = 5

= \frac{( 2 2 ) ^{2} - ( 5 ) ^{2}}{2 2}

\frac{( 2 2 ) ^{2} - ( 5 ) ^{2}}{2 2} = \frac{3}{2 2}

= \frac{3}{2 2}

18 x^{2} + 4 y^{2} - 72 x + 4 y + 37 = 0

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

\frac{x ^{2}}{12} + \frac{y ^{2}}{49} = 1

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

\frac{x ^{2}}{169} + \frac{y ^{2}}{25} = 1

What is the eccentricity (x^2)/5+(y^2)/8 =1 ?
The eccentricity (x^2)/5+(y^2)/8 =1 is (sqrt(3))/(2sqrt(2))