An ellipse frac{x^2}{a^2}+frac{y^2}{b^2}=1 (a | vedclass

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(B) Let the rectangle be $ABCD$ with dimensions $2a$ and $2b$. The center of the ellipse is the origin $O(0,0)$. The vertices of the rectangle are $(a

Vedclass · Accepted Answer

$\frac{1}{2}$

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(B) Let the rectangle be $ABCD$ with dimensions $2a$ and $2b$. The center of the ellipse is the origin $O(0,0)$. The vertices of the rectangle are $(a, b), (-a, b), (-a, -b),$ and $(a, -b)$.Let the angle between the diagonals be $2	heta$. From the geometry of the rectangle,$	an 	heta = \frac{b}{a}$.The angle between the diagonals is given as $	an(2	heta) = 4\sqrt{3}$.Using the formula $	an(2	heta) = \frac{2	an 	heta}{1-	an^2 	heta}$,we have:$4\sqrt{3} = \frac{2(b/a)}{1-(b/a)^2}$$2\sqrt{3} = \frac{b/a}{1-(b/a)^2}$Let $k = b/a$. Then $2\sqrt{3}(1-k^2) = k$,which implies $2\sqrt{3}k^2 + k - 2\sqrt{3} = 0$.Solving the quadratic equation for $k$:$k = \frac{-1 \pm \sqrt{1 - 4(2\sqrt{3})(-2\sqrt{3})}}{2(2\sqrt{3})} = \frac{-1 \pm \sqrt{1 + 48}}{4\sqrt{3}} = \frac{-1 \pm 7}{4\sqrt{3}}$.Since $k = b/a > 0$,we take $k = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.The eccentricity $e$ of the ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - k^2}$.$e = \sqrt{1 - (\frac{\sqrt{3}}{2})^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,the correct option is $(b)$.

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ is inscribed in a rectangle of dimensions $2a$ and $2b$ respectively. If the angle between the diagonals of the rectangle is $\tan^{-1}(4\sqrt{3})$,then the eccentricity of that ellipse is

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