What is the eccentricity (x^2)/(324)+(y^2)/(400)=1 ?

The eccentricity (x^2)/(324)+(y^2)/(400)=1 is (sqrt(19))/(10)

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eccentricity (x^2)/(324)+(y^2)/(400)=1

eccentricity

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

\frac{19}{10}

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 20, a = 18

= \frac{2 0 ^{2} - 1 8 ^{2}}{20}

\frac{2 0 ^{2} - 1 8 ^{2}}{20} = \frac{19}{10}

= \frac{19}{10}

\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

3 x^{2} + y^{2} - 24 x = - 39

9 x^{2} + 25 y^{2} + 18 x + 150 y + 9 = 0

What is the eccentricity (x^2)/(324)+(y^2)/(400)=1 ?
The eccentricity (x^2)/(324)+(y^2)/(400)=1 is (sqrt(19))/(10)