Let x=4 be a directrix to an ellipse whose ce | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Given the directrix $x = \frac{a}{e} = 4$ and eccentricity $e = \frac{1}{2

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$4x-2y=1$

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(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Given the directrix $x = \frac{a}{e} = 4$ and eccentricity $e = \frac{1}{2}$,we find $a = 4 	imes \frac{1}{2} = 2$. Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$. Thus,the ellipse equation is $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Since $P(1, \beta)$ lies on the ellipse,$\frac{1^2}{4} + \frac{\beta^2}{3} = 1 \Rightarrow \frac{\beta^2}{3} = \frac{3}{4} \Rightarrow \beta^2 = \frac{9}{4} \Rightarrow \beta = \frac{3}{2}$ (as $\beta > 0$). The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Substituting $a^2=4, b^2=3, x_1=1, y_1=\frac{3}{2}$,we get $\frac{4x}{1} - \frac{3y}{3/2} = 4 - 3$. $4x - 2y = 1$.

Let $x=4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$. If $P(1, \beta), \beta>0$ is a point on this ellipse,then the equation of the normal to it at $P$ is

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