Gujarati
Hindi
10-2. Parabola, Ellipse, Hyperbola
normal

The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is

A

$\frac{{\sqrt 3 }}{2}$

B

$\frac {1}{9}$

C

$\frac{1}{{\sqrt 3 }}$

D

$\frac {1}{3}$

Solution

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

$9(x-3)^{2}+8 y^{2}-72 y+144=0$

$9(x-3)^{2}+8\left[\left(y-\frac{9}{2}\right)^{2}-\frac{81}{4}\right]+144=0$

$9(x-3)^{2}+8\left(y-\frac{9}{2}\right)^{2}=18$

$\frac{(x-3)^{2}}{2}+\frac{(y-9 / 2)}{9 / 4}=1$

$ \Rightarrow {e^2} = 1 – \frac{{24}}{9} = \frac{1}{9}$

$\mathrm{e}=\frac{1}{3}$

Standard 11
Mathematics

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