Gujarati
Hindi
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

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