The eccentricity of an ellipse whose centre i | vedclass

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(C) Given eccentricity $e = \frac{1}{2}$.The equation of the directrix is $x = -\frac{a}{e} = -4$,so $\frac{a}{1/2} = 4$,which gives $a = 2$.We know $

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$4x - 2y = 1$

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(C) Given eccentricity $e = \frac{1}{2}$.The equation of the directrix is $x = -\frac{a}{e} = -4$,so $\frac{a}{1/2} = 4$,which gives $a = 2$.We know $b^2 = a^2(1 - e^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{3} = 1$.Differentiating with respect to $x$,we get $\frac{2x}{4} + \frac{2y}{3} \frac{dy}{dx} = 0$,which simplifies to $\frac{dy}{dx} = -\frac{3x}{4y}$.At the point $\left(1, \frac{3}{2}\right)$,the slope of the tangent is $m_t = -\frac{3(1)}{4(3/2)} = -\frac{3}{6} = -\frac{1}{2}$.The slope of the normal is $m_n = -\frac{1}{m_t} = 2$.The equation of the normal at $\left(1, \frac{3}{2}\right)$ is $y - \frac{3}{2} = 2(x - 1)$.Multiplying by $2$,we get $2y - 3 = 4x - 4$,which simplifies to $4x - 2y = 1$.

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

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