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

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

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eccentricity (x^2)/3+(y^2)/5 =1

eccentricity

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

\frac{2}{5}

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 3

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

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

= \frac{2}{5}

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

14 x^{2} + 9 y^{2} = 126

10 x^{2} + 3 y^{2} = 30

\frac{y ^{2}}{64} + \frac{x ^{2}}{36} = 1

3 x^{2} + y^{2} + 2 x - 6 y = 3

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