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10-2. Parabola, Ellipse, Hyperbola
normal
The eccentricity of the ellipse $ (x - 3)^2 + (y - 4)^2 =$ $\frac{{{y^2}}}{9}\,$ is
A
$\frac{{\sqrt 3 }}{2}\,$
B
$\frac{1}{3}\,$
C
$\frac{1}{{3\sqrt 2 }}\,$
D
$\frac{1}{{\sqrt 3 \,}}\,$
Solution
$9(x – 3)^2 + 9(y – 4)^2 = y^2$
$9(x – 3)^2 + 8y^2 – 72y + 14y = 0$
$9(x – 3)^2 + 8(y^2 – 9y) + 144 = 0$
$9(x – 3)^2 + 8 $ $\left[ {{{\left( {y – \frac{9}{2}} \right)}^2} – \frac{{81}}{4}} \right]$ $ + 144 = 0$
$\Rightarrow$ $9(x – 3)^2 +$ $8{\left( {y – \frac{9}{2}} \right)^2}$ $= 162 – 144 = 18$
$\frac{{9{{(x – 3)}^2}}}{{18}} + \frac{{8\left( {y – \frac{9}{2}} \right)}}{{18}} = 1$
$\Rightarrow$ $\frac{{{{(x – 3)}^2}}}{2} + \frac{{\left( {y – \frac{9}{2}} \right)}}{{\frac{9}{4}}} = 1$
$e^2 = 1 -$ $\frac{{2\,\cdot\,4}}{9}$ $=$ $\frac{1}{9}$ ;
$\therefore$ $e =$ $\frac{1}{3}$
Standard 11
Mathematics
