What is the eccentricity ((x-1)^2)/6+((y-3)^2)/2 =1 ?

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

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eccentricity ((x-1)^2)/6+((y-3)^2)/2 =1

eccentricity

\frac{( x - 1 ) ^{2}}{6} + \frac{( y - 3 ) ^{2}}{2} = 1

\frac{2}{3}

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{6} + \frac{( y - 3 ) ^{2}}{2} = 1 :

Ellipse with center

(h, k) = (1, 3), a = 6, b = 2

= \frac{( 6 ) ^{2} - ( 2 ) ^{2}}{6}

\frac{( 6 ) ^{2} - ( 2 ) ^{2}}{6} = \frac{2}{3}

= \frac{2}{3}

25 x^{2} + 9 y^{2} - 100 x + 18 y - 116 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{36} = 1

\frac{x ^{2}}{8} + \frac{y ^{2}}{3} = 1

10 x^{2} + 5 y^{2} = 50

9 x^{2} + 16 y^{2} - 144 = 0

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