Q68 / 125 · BITSAT 2015 · Mathematics

The eccentricity of an ellipse, with its centre at the origin, is $1 / 2$. If one of the directrices is $\mathrm{x}=4$ then the equation of the ellipse is:

  • A
    $4 x^{2}+3 y^{2}=1$
  • Correct answer
    B
    $3 x^{2}+4 y^{2}=12$
  • C
    $4 x^{2}+3 y^{2}=12$
  • D
    $3 x^{2}+4 y^{2}=1$

Solution

$\mathrm{e}=\frac{1}{2} .$ Directrix $, \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4$ $$ \therefore \mathrm{a}=4 \times \frac{1}{2}=2 \quad \therefore \mathrm{b}=2 \sqrt{1-\frac{1}{4}}=\sqrt{3} $$ Equation of ellispe is $$ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 \Rightarrow 3 x^{2}+4 y^{2}=12 $$

Hence, the correct answer is option: 'B'.

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