A) \[9{{x}^{2}}+4{{y}^{2}}=36\]
B) \[4{{x}^{2}}+9{{y}^{2}}=36\]
C) \[36{{x}^{2}}+9{{y}^{2}}=4\]
D) \[9{{x}^{2}}+36{{y}^{2}}=4\]
Correct Answer: B
Solution :
Given, eccentricity \[e=\frac{\sqrt{5}}{3}\] and foci \[=(\pm \sqrt{5},\,0)\] \[\Rightarrow \] \[ae=\sqrt{5}\] \[\Rightarrow \] \[a=\frac{\sqrt{5}\times 3}{\sqrt{5}}=3\] Now, \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})=9\left( 1-\frac{5}{9} \right)\] \[\Rightarrow \] \[{{b}^{2}}=4\] \[\therefore \] The equation of ellipse is \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1\] \[\Rightarrow \] \[4{{x}^{2}}+9{{y}^{2}}=36\]You need to login to perform this action.
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