A) \[2/5\]
B) \[4/5\]
C) \[1/3\]
D) \[1/5\]
Correct Answer: B
Solution :
Key Idea: If equation of ellipse is \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] when, \[a>b,\] \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}\] when,\[a<b\] \[e=\sqrt{1-\frac{{{a}^{2}}}{{{b}^{2}}}}\] Given equation is \[9{{x}^{2}}+25{{y}^{2}}=225\] or it can be written as \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\] On comparing with\[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\] We get \[{{a}^{2}}=25,\,\,{{b}^{2}}=9\] \[\therefore \] \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{9}{25}}\] \[=\sqrt{\frac{16}{25}}=\frac{4}{5}\] Note: Eccentricity of ellipse lies between zero to one.
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