A) \[3{{x}^{2}}+4{{y}^{2}}=1\]
B) \[4{{x}^{2}}+3{{y}^{2}}=12\]
C) \[3{{x}^{2}}+4{{y}^{2}}=12\]
D) \[4{{x}^{2}}+3{{y}^{2}}=12\]
Correct Answer: C
Solution :
Given, eccentricity, \[e=\frac{1}{2}\] \[\Rightarrow \] \[\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\frac{1}{2}\] \[\Rightarrow \] \[1-\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{1}{4}\] \[\Rightarrow \] \[\frac{{{b}^{2}}}{{{a}^{2}}}=1-\frac{1}{4}\] \[=\frac{3}{4}\] \[\Rightarrow \] \[{{b}^{2}}=\frac{3}{4}{{a}^{2}}\] ?..(i) Also, given equation of one of its direction is \[x=4\], \[\therefore \] \[\frac{a}{e}=4\] \[\Rightarrow \] \[a=4e=4\times \frac{1}{2}=2\] From Eq. (i), we get \[{{b}^{2}}=\frac{3}{4}{{(2)}^{2}}=3\] Hence, required equation of ellipse is \[\frac{{{x}^{2}}}{{{(2)}^{2}}}+\frac{{{y}^{2}}}{3}=1\] \[\Rightarrow \] \[3{{x}^{2}}+4{{y}^{2}}=12\]
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