A) 4
B) \[2\sqrt{2}\]
C) \[4\sqrt{2}\]
D) 2
Correct Answer: A
Solution :
\[{{m}_{SB}}.\,{{m}_{S'B}}=-1\] |
\[{{b}^{2}}={{a}^{2}}{{e}^{2}}\] |
\[\frac{1}{2}S'B.SB=8\] |
\[{{a}^{2}}{{e}^{2}}+{{b}^{2}}=16\] |
\[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] |
Using (i), (ii), (iii), a = 4 |
\[b=2\sqrt{2}\] |
\[e=\frac{1}{\sqrt{2}}\] |
\[\therefore \] \[\ell (L.R.)=\frac{2{{b}^{2}}}{a}=4.\] |
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