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The eccentricity of an ellipse whose centre is at the origin is $${1 \over 2}$$. If one of its directrices is x = – 4, then the equation of the normal to it at $$\left( {1,{3 \over 2}} \right)$$ is :

JEE · Math · previous-year question

  1. A.2y – x = 2
  2. B.4x – 2y = 1correct
  3. C.4x + 2y = 7
  4. D.x + 2y = 4

Answer

B. 4x – 2y = 1

Explanation

Given e = $${1 \over 2}$$ and $${a \over e}$$ = 4 $$ \therefore $$ $$a$$ = 2 We have b2 = $$a$$2 (1 – e2) = $$4\left( {1 - {1 \over 4}} \right)$$ = 3 $$ \therefore $$ Equation of ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$ Now, the equation of normal at $$\left( {1,{3 \over 2}} \right)$$ is $${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$$ $$ \Rightarrow $$ $${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$$ $$ \Rightarrow $$ 4x – 2y = 1

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