What is the eccentricity (x^2)/9+(y^2)/(16)>1 ?

The eccentricity (x^2)/9+(y^2)/(16)>1 is (sqrt(7))/4

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eccentricity (x^2)/9+(y^2)/(16)>1

eccentricity

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

\frac{7}{4}

Solution steps

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

Rewrite as

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 4, a = 3

= \frac{4 ^{2} - 3 ^{2}}{4}

\frac{4 ^{2} - 3 ^{2}}{4} = \frac{7}{4}

= \frac{7}{4}

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952

3 x^{2} + y^{2} - 24 x + 39 = 0

What is the eccentricity (x^2)/9+(y^2)/(16)>1 ?
The eccentricity (x^2)/9+(y^2)/(16)>1 is (sqrt(7))/4