L_1^{prime} is the end of a latus rectum of t | vedclass

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(B) The given equation of the ellipse is $3x^2 + 4y^2 = 12$,which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here,$a^2 = 4$ and $b^2 = 3$,

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$\frac{11}{19}$

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(B) The given equation of the ellipse is $3x^2 + 4y^2 = 12$,which can be written as $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Here,$a^2 = 4$ and $b^2 = 3$,so $a = 2$ and $b = \sqrt{3}$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$. The coordinates of the end of the latus rectum in the third quadrant are $(-ae, -\frac{b^2}{a}) = (-2 	imes \frac{1}{2}, -\frac{3}{2}) = (-1, -\frac{3}{2})$. The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Substituting $(x_1, y_1) = (-1, -\frac{3}{2})$,$a^2 = 4$,and $b^2 = 3$: $\frac{4x}{-1} - \frac{3y}{-3/2} = 4 - 3$ $\Rightarrow -4x + 2y = 1$ $\Rightarrow y = 2x + \frac{1}{2}$. Substituting $y = 2x + \frac{1}{2}$ into the ellipse equation $3x^2 + 4y^2 = 12$: $3x^2 + 4(2x + \frac{1}{2})^2 = 12$ $\Rightarrow 3x^2 + 4(4x^2 + 2x + \frac{1}{4}) = 12$ $\Rightarrow 3x^2 + 16x^2 + 8x + 1 = 12$. $19x^2 + 8x - 11 = 0$. Factoring the quadratic: $(x + 1)(19x - 11) = 0$. The roots are $x = -1$ (the point $L_1^{\prime}$) and $x = \frac{11}{19}$ (the point $P$). Thus,$a = \frac{11}{19}$.

$L_1^{\prime}$ is the end of a latus rectum of the ellipse $3x^2 + 4y^2 = 12$ which is lying in the third quadrant. If the normal drawn at $L_1^{\prime}$ to this ellipse intersects the ellipse again at the point $P(a, b)$,then $a =$

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