On the ellipse frac{x^{2}}{8}+frac{y^{2}}{4}= | vedclass

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(B) The equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$. Here $a^{2}=8$ and $b^{2}=4$.The eccentricity $e$ is given by $e = \sqrt{1 - \

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$6$

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(B) The equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$. Here $a^{2}=8$ and $b^{2}=4$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{4}{8}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.The foci are $S(-ae, 0)$ and $S'(ae, 0)$,where $ae = \sqrt{8} \cdot \frac{1}{\sqrt{2}} = 2$. So,$S(-2, 0)$ and $S'(2, 0)$.The line perpendicular to $x+2y=0$ has the form $2x-y+k=0$,or $y=2x+k$.The condition for tangency $y=mx+c$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}+b^{2}$.Here $m=2$,so $c^{2} = 8(2^{2}) + 4 = 32+4 = 36$,which gives $c = \pm 6$.Since $P$ is in the second quadrant,the tangent must have a positive $y$-intercept,so $y=2x+6$.The point of tangency $P(x_{1}, y_{1})$ is given by $(-\frac{a^{2}m}{c}, \frac{b^{2}}{c}) = (-\frac{8 \cdot 2}{6}, \frac{4}{6}) = (-\frac{8}{3}, \frac{2}{3})$.The area $A$ of $\Delta SPS'$ with base $SS'=4$ and height $|y_{1}| = \frac{2}{3}$ is $A = \frac{1}{2} \cdot 4 \cdot \frac{2}{3} = \frac{4}{3}$.Finally,$(5-e^{2}) \cdot A = (5 - \frac{1}{2}) \cdot \frac{4}{3} = \frac{9}{2} \cdot \frac{4}{3} = 6$.

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