The eccentricity of the ellipse with minor ax | vedclass

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The foci are $S(ae, 0)$ and $S'(-ae, 0)$.The upper vertex

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$\sin \alpha$

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The foci are $S(ae, 0)$ and $S'(-ae, 0)$.The upper vertex is $V(0, b)$.The line segment joining the foci is $SS'$,which subtends an angle $2\alpha$ at $V(0, b)$.This means $\angle SVS' = 2\alpha$,so $\angle SVO = \alpha$,where $O$ is the origin $(0, 0)$.In the right-angled triangle $	riangle SVO$,we have $	an \alpha = \frac{SO}{VO} = \frac{ae}{b}$.Thus,$ae = b 	an \alpha$.We know the relation $b^2 = a^2(1 - e^2)$,which implies $b^2 = a^2 - a^2e^2$.Substituting $ae = b 	an \alpha$,we get $b^2 = a^2 - (b 	an \alpha)^2$,so $a^2 = b^2 + b^2 	an^2 \alpha = b^2(1 + 	an^2 \alpha) = b^2 \sec^2 \alpha$.Therefore,$a = b \sec \alpha$.Now,the eccentricity $e = \frac{ae}{a} = \frac{b 	an \alpha}{b \sec \alpha} = \frac{\sin \alpha / \cos \alpha}{1 / \cos \alpha} = \sin \alpha$.

The eccentricity of the ellipse with minor axis $2b$,if the line segment joining the foci subtends an angle $2\alpha$ at the upper vertex,is equal to

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