Find the eccentricity of the ellipse $(x-3)^2+(y-4)^2=\frac{y^2}{9}$

Question

Find the eccentricity of the ellipse $(x-3)^2+(y-4)^2=\frac{y^2}{9}$

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$(x-3)^2+(y-4)^2=\frac{y^2}{9}$

$x^2-6x+9+y^2-8y+16-\frac{y^2}{9}=0$

$(x-3)^2+\frac{8y^2}{9}-8y+16=0$

$(x-3)^2+\frac{8}{9}(y^2-9y)+16=0$

$(x-3)^2+\frac{8}{9}(y^2-9y+9)+8=0$

$(x-3)^2+\frac{8}{9}(y-3)^2+8=0$

$\frac{(x-3)^2}{8}+\frac{(y-3)^2}{9}=-1$

But this equation does not appear to be in the standard form,$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.It does not even appear ro be the equation of the ellipse .But in the book its eccentricity is given as $\frac{1}{3}$.Have i done somewhere wrong?If yes,what is the correct way to find its eccentricity?

Answer 1

Please check step no 5 . $(y-3)^2$ is not equal $y^2-9y+9$ and then proceed.