A) \[4{{x}^{2}}+3{{y}^{2}}=1\]
B) \[3{{x}^{2}}+4{{y}^{2}}=12\]
C) \[4{{x}^{2}}+3{{y}^{2}}=12\]
D) \[3{{x}^{2}}+4{{y}^{2}}=1\]
Correct Answer: B
Solution :
Since directrix is parallel to y-axis, hence axes of the ellipse are parallel to x-axis. Let the equation of the ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], \[(a>b)\] \[{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}\Rightarrow \frac{{{b}^{2}}}{{{a}^{2}}}=1-{{e}^{2}}=1-\frac{1}{4}\Rightarrow \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{3}{4}\]. Also, one of the directrices is \[x=4\] \[\Rightarrow \]\[\frac{a}{e}=4\Rightarrow a=4e=4.\frac{1}{2}=2\]; \[{{b}^{2}}=\frac{3}{4}{{a}^{2}}=\frac{3}{4}.4=3\] \[\therefore \] Required ellipse is \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\] or \[3{{x}^{2}}+4{{y}^{2}}=12\].
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