A) 2
B) 4
C) \[4\sqrt{2}\]
D) \[2\sqrt{2}\]
Correct Answer: B
Solution :
Since \[\angle S'BS={{90}^{o}}\] \[\therefore \]Slope of \[S'B\times \]Slope of \[SB=-1\] \[\Rightarrow \]\[\frac{-b}{-ae}\times \frac{-b}{ae}=-1\Rightarrow {{b}^{2}}={{a}^{2}}{{e}^{2}}\] ?(i) Now, area of \[\Delta S'BS=8\] \[\Rightarrow \]\[\frac{1}{2}\times 2ae\times b=8\] \[\Rightarrow \]\[ae\times b=8\] \[\Rightarrow \]\[{{a}^{2}}{{e}^{2}}\times {{b}^{2}}=64\] \[\Rightarrow \]\[{{({{a}^{2}}{{e}^{2}})}^{2}}=64\] [Using (i)] \[\Rightarrow \]\[{{a}^{2}}{{e}^{2}}=8\Rightarrow ae=\pm 2\sqrt{2}\] \[\therefore \]\[b=2\sqrt{2}\] Again,\[{{a}^{2}}={{a}^{2}}{{e}^{2}}+{{b}^{2}}\] \[\left[ \because e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}} \right]\] \[\Rightarrow \]\[{{a}^{2}}=8+8=16\Rightarrow a=\pm 4\] \[\therefore \]Length of latus rectum\[=\frac{2{{b}^{2}}}{a}=\frac{2\times 8}{4}=4\]You need to login to perform this action.
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