For an ellipse with eccentricity $\frac{1}{2}$ the centre is at the origin. If one directrix is $x=4$, then…

For an ellipse with eccentricity $\frac{1}{2}$ the centre is at the origin. If one directrix is $x=4$, then the equation of the ellipse is
  1. $3 x^2+4 y^2=1$
  2. $3 x^2+4 y^2=12$
  3. $4 x^2+3 y^2=1$
  4. $4 x^2+3 y^2=12$

Solution

Given that, $e=\frac{1}{2}$ and $\frac{a}{e}=4$ $ \begin{array}{lrl} \Rightarrow & \frac{a}{1 / 2}=4 \Rightarrow a=2 \\ \text { Using } & b^2=a^2\left(1-e^2\right) \\ \Rightarrow & b^2=4\left(1-\frac{1}{4}\right)=3 \end{array} $ $\therefore$ Equation of ellipse is $ \begin{aligned} \frac{x^2}{4}+\frac{y^2}{3} & =1 \\ \Rightarrow \quad 3 x^2+4 y^2 & =12 \end{aligned} $

Asked in: AP EAMCET 2008

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