What is the eccentricity (x^2)/(196)+(y^2)/(121)=1 ?

The eccentricity (x^2)/(196)+(y^2)/(121)=1 is (5sqrt(3))/(14)

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eccentricity (x^2)/(196)+(y^2)/(121)=1

eccentricity

\frac{x ^{2}}{196} + \frac{y ^{2}}{121} = 1

\frac{5 3}{14}

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{196} + \frac{y ^{2}}{121} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 14, b = 11

= \frac{1 4 ^{2} - 1 1 ^{2}}{14}

\frac{1 4 ^{2} - 1 1 ^{2}}{14} = \frac{5 3}{14}

= \frac{5 3}{14}

\frac{x ^{2}}{5} + \frac{y ^{2}}{20.98} = 1

16 x^{2} + 192 x + 576 + 25 y^{2} = 1600

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

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

4 x^{2} + 9 y^{2} = 144

What is the eccentricity (x^2)/(196)+(y^2)/(121)=1 ?
The eccentricity (x^2)/(196)+(y^2)/(121)=1 is (5sqrt(3))/(14)