The tangents at the extremities of the latus | vedclass

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(D) 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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$16$

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(D) 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 $b^2 = 3$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$.The coordinates of the extremities of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 1, \pm \frac{3}{2})$.The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$. For $(1, \frac{3}{2})$,the tangent is $\frac{x(1)}{4} + \frac{y(3/2)}{3} = 1$,which simplifies to $\frac{x}{4} + \frac{y}{2} = 1$,or $x + 2y = 4$.Due to symmetry,the four tangents are $x + 2y = 4$,$x - 2y = 4$,$-x + 2y = 4$,and $-x - 2y = 4$.These lines form a rhombus with vertices at $(4, 0)$,$(0, 2)$,$(-4, 0)$,and $(0, -2)$.The area of the rhombus is $\frac{1}{2} 	imes 	ext{diagonal}_1 	imes 	ext{diagonal}_2 = \frac{1}{2} 	imes 8 	imes 4 = 16 \ sq. \ units$.

The tangents at the extremities of the latus rectum of the ellipse $3x^2 + 4y^2 = 12$ form a rhombus of area (in $sq. \ units$):

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