Let S and S' be the foci of an ellipse and B | vedclass

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

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$4$

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $S(ae, 0)$ and $S'(-ae, 0)$,and the extremity of the minor axis is $B(0, b)$.Since $\Delta S'BS$ is a right-angled triangle at $B$,the product of the slopes of $BS$ and $BS'$ is $-1$.Slope of $BS = \frac{b-0}{0-ae} = -\frac{b}{ae}$.Slope of $BS' = \frac{b-0}{0-(-ae)} = \frac{b}{ae}$.Since the angle at $B$ is $90^\circ$,the product of slopes is $-1$,so $(-\frac{b}{ae}) 	imes (\frac{b}{ae}) = -1$,which implies $b^2 = a^2e^2$.We know that for an ellipse,$b^2 = a^2(1-e^2)$,so $a^2e^2 = a^2 - a^2e^2$,which gives $a^2e^2 = \frac{a^2}{2}$,so $b^2 = \frac{a^2}{2}$.The area of $\Delta S'BS = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2ae) 	imes b = aeb = 8$.Squaring both sides,$a^2e^2b^2 = 64$. Substituting $a^2e^2 = b^2$,we get $b^4 = 64$,so $b^2 = 8$.Since $b^2 = \frac{a^2}{2}$,we have $8 = \frac{a^2}{2}$,so $a^2 = 16$,which means $a = 4$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2(8)}{4} = 4$.

Let $S$ and $S'$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta S'BS$ is a right-angled triangle with the right angle at $B$ and $\text{Area}(\Delta S'BS) = 8 \text{ sq. units}$,then the length of the latus rectum of the ellipse is

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