What is the eccentricity (x^2}{16}+\frac{y^2)/7 =1 ?

The eccentricity (x^2}{16}+\frac{y^2)/7 =1 is 3/4

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eccentricity (x^2}{16}+\frac{y^2)/7 =1

eccentricity

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

\frac{3}{4}

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 7

= \frac{4 ^{2} - ( 7 ) ^{2}}{4}

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

= \frac{3}{4}

(\frac{x - 1}{4})^{2} + (\frac{y + 2}{3})^{2} = 1

4 x^{2} + 9 y^{2} - 8 x - 36 y + 4 = 0

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

9 x^{2} + 36 y^{2} = 324

5 x^{2} + 7 y^{2} = 35

What is the eccentricity (x^2}{16}+\frac{y^2)/7 =1 ?
The eccentricity (x^2}{16}+\frac{y^2)/7 =1 is 3/4