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Q.

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

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a

$x + 2 y = 4$

b

$2 x - y = 2$

c

$4 x + 2 y = 7$

d

$4 x - 2 y = 1$

answer is C.

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Detailed Solution

$e = 1 / 2 & x = - a / e = - 4 \Rightarrow a = 2$
$b^{2} = a^{2} (1 - e^{2}) = 4 (1 - 1 / 4) = 3 Eq of ellipse = \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$
Eq of normal at $(1, 3 / 2) is \frac{4 x}{1} - \frac{3 y}{3 / 2} = 4 - 3 \Rightarrow 4 x - 2 y = 1$

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