• # question_answer The eccentricity of an ellipse, with its centre at the origin, is $\frac{1}{2}$. If one of the directrices is $x=4$, then the equation of the ellipse is [AIEEE 2004] 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)            $3{{x}^{2}}+4{{y}^{2}}=1$

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 $3{{x}^{2}}+4{{y}^{2}}=12$.