The eccentricity of an ellipse, with its centre as origin, is \(1 / 2\). If one of the directrices is…

The eccentricity of an ellipse, with its centre as origin, is \(1 / 2\). If one of the directrices is \(x=4\), then the equation of the ellipse is given by
  1. \(4 x^2+y^2=12\)
  2. \(x^2+3 y^2=12\)
  3. \(4 x^2+3 y^2=12\)
  4. \(3 x^2+4 y^2=12\)

Solution

Centre \(=(0,0)\) Eccentricity \((\mathrm{e})=\frac{1}{2}\) Equation of directrix is \(x=4\) \(\Rightarrow \frac{a}{e}=4\) \(\Rightarrow a=4 e\) \(\Rightarrow a=2\) \(\begin{aligned} b^2 & =a^2\left(1-e^2\right) \\ & =4\left(1-\frac{1}{4}\right) \\ b^2 & =3 \end{aligned}\) \(\therefore\) Required Equation of Ellipse is \(\frac{x^2}{4}+\frac{y^2}{3}=1\) \(3 x^2+4 y^2=12\) Hence, option (d) is correct.

Asked in: AP EAMCET 2020 (18 Sep Shift 2)

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