A) \[5/6\]
B) \[3/5\]
C) \[\sqrt{2}/3\]
D) \[\sqrt{5}/3\]
Correct Answer: D
Solution :
Key Idea: Let the equation of ellipse be\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] When\[a>b,\,\,e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}\]and when\[b>a,\,\,e=\sqrt{1-\frac{{{a}^{2}}}{{{b}^{2}}}}\] Given equation is \[4{{x}^{2}}+9{{y}^{2}}+8x+36y+4=0\] \[\Rightarrow \] \[4({{x}^{2}}+2x+1)+9({{y}^{2}}+4y+4)+4\] \[4+36\] \[\Rightarrow \] \[4{{(x+1)}^{2}}+9{{(y+2)}^{2}}=36\] \[\Rightarrow \] \[\frac{{{(x+1)}^{2}}}{9}+\frac{{{(y+2)}^{2}}}{4}=1\] \[\therefore \] \[e=\sqrt{1-\frac{4}{9}}\] \[(\because \,\,a>b)\] \[=\frac{\sqrt{5}}{3}\] Note: Eccentricity cannot be negative or imaginary.
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