The eccentricity of the ellipse (x - 3)^2 + | vedclass

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Question

$9(x - 3)^2 + 9(y - 4)^2 = y^2$

$9(x - 3)^2 + 8y^2 - 72y + 14y = 0$

$9(x - 3)^2 + 8(y^2 - 9y) + 144 = 0$

$9(x - 3)^2 + 8 $ $\left[ {{{\l

Vedclass · Accepted Answer

$\frac{1}{3}\,$

Vedclass · Answer

$9(x - 3)^2 + 9(y - 4)^2 = y^2$

$9(x - 3)^2 + 8y^2 - 72y + 14y = 0$

$9(x - 3)^2 + 8(y^2 - 9y) + 144 = 0$

$9(x - 3)^2 + 8 $ $\left[ {{{\left( {y - \frac{9}{2}} ight)}^2} - \frac{{81}}{4}} ight]$ $ + 144 = 0$

$\Rightarrow$ $9(x - 3)^2 +$ $8{\left( {y - \frac{9}{2}} ight)^2}$ $= 162 - 144 = 18$

$\frac{{9{{(x - 3)}^2}}}{{18}} + \frac{{8\left( {y - \frac{9}{2}} ight)}}{{18}} = 1$

$\Rightarrow$ $\frac{{{{(x - 3)}^2}}}{2} + \frac{{\left( {y - \frac{9}{2}} ight)}}{{\frac{9}{4}}} = 1$ 

$e^2 = 1 -$ $\frac{{2\,\cdot\,4}}{9}$ $=$ $\frac{1}{9}$ ;

$	herefore$ $e =$ $\frac{1}{3}$

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

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