What is the eccentricity (x^2)/4+(y^2)/(25)=1 ?

The eccentricity (x^2)/4+(y^2)/(25)=1 is (sqrt(21))/5

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eccentricity (x^2)/4+(y^2)/(25)=1

eccentricity

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

\frac{21}{5}

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 2

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

\frac{5 ^{2} - 2 ^{2}}{5} = \frac{21}{5}

= \frac{21}{5}

9 x^{2} + 49 y^{2} = 441

\frac{x ^{2}}{324} + \frac{y ^{2}}{400} = 1

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

4 x^{2} + 3 y^{2} + 24 x + 36 y + 72 = 0

\frac{5}{81} x^{2} + \frac{y ^{2}}{9} = 1

What is the eccentricity (x^2)/4+(y^2)/(25)=1 ?
The eccentricity (x^2)/4+(y^2)/(25)=1 is (sqrt(21))/5