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  4. The eccentricity of the ellipse 25x2+9y2-150x-90y+225=0 is

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Q. The eccentricity of the ellipse $ 25{{x}^{2}}+9{{y}^{2}}-150x-90y+225=0 $ is

J & K CETJ & K CET 2013

Solution:

Given curve is $ 25{{x}^{2}}+9{{y}^{2}}-150x-90y+225=0 $
$ \Rightarrow $ $ (25{{x}^{2}}-150x)+(9{{y}^{2}}-90y)+225=0 $
$ \Rightarrow $ $ 25({{x}^{2}}-6x)+9({{y}^{2}}-10y)+225=0 $
$ \Rightarrow $ $ 25\{{{x}^{2}}-6x+9-9\}+9\{{{y}^{2}}-10y+25-25\} $
$ +225=0 $
$ \Rightarrow $ $ 25{{(x-3)}^{2}}-225+9{{(y-5)}^{2}}-225+225=0 $
$ \Rightarrow $ $ 25{{(x-3)}^{2}}+9{{(y-5)}^{2}}=225 $
$ \Rightarrow $ $ \frac{{{(x-3)}^{2}}}{9}+\frac{{{(y-5)}^{2}}}{25}=1 $
$ [\because $ $ (b>a) $ ]
Which represent an ellipse Whose eccentricity
$ e=\sqrt{\frac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} $