The area of the quadrilateral formed by drawi | vedclass

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(A) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the ends of the latus recta are $(\pm ae, \pm \frac{b^2}{a})$.Given $a^2 = 4$ and $b^2 = 1

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$\frac{16}{\sqrt{3}}$

Vedclass · Answer

(A) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the ends of the latus recta are $(\pm ae, \pm \frac{b^2}{a})$.Given $a^2 = 4$ and $b^2 = 1$,we have $a = 2$ and $b = 1$.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.The coordinates of the ends of the latus recta are $(\pm \sqrt{3}, \pm \frac{1}{2})$.The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.For $(\sqrt{3}, \frac{1}{2})$,the tangent is $\frac{x\sqrt{3}}{4} + \frac{y(1/2)}{1} = 1 \Rightarrow \frac{\sqrt{3}}{4}x + \frac{1}{2}y = 1$.The four tangents form a rhombus with vertices at $(\pm \frac{a^2}{ae}, 0) = (\pm \frac{4}{\sqrt{3}}, 0)$ and $(0, \pm \frac{b^2}{b^2/a}) = (0, \pm a) = (0, \pm 2)$.The area of the rhombus is $2 	imes 	ext{base} 	imes 	ext{height} = 2 	imes |\frac{4}{\sqrt{3}}| 	imes |2| = \frac{16}{\sqrt{3}}$.

The area of the quadrilateral formed by drawing tangents at the ends of the latus recta of the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ is

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