What is the eccentricity (x^2)/(25)+((y+3)^2)/(49)=1 ?

The eccentricity (x^2)/(25)+((y+3)^2)/(49)=1 is (2sqrt(6))/7

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eccentricity (x^2)/(25)+((y+3)^2)/(49)=1

eccentricity

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1

\frac{2 6}{7}

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, - 3), b = 7, a = 5

= \frac{7 ^{2} - 5 ^{2}}{7}

\frac{7 ^{2} - 5 ^{2}}{7} = \frac{2 6}{7}

= \frac{2 6}{7}

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

9 x^{2} + 4 y^{2} - 54 x + 32 y + 1 = 0

16 x^{2} + 25 y^{2} - 96 x + 100 y - 156 = 0

12 x^{2} + 8 y^{2} = 96

14 x^{2} + 8 y^{2} = 112

What is the eccentricity (x^2)/(25)+((y+3)^2)/(49)=1 ?
The eccentricity (x^2)/(25)+((y+3)^2)/(49)=1 is (2sqrt(6))/7