A) \[\frac{5}{\sqrt{6}}\]
B) \[\frac{10}{\sqrt{6}}\]
C) \[\frac{8}{\sqrt{6}}\]
D) None of these
Correct Answer: B
Solution :
Here, \[{{a}^{2}}+2>{{a}^{2}}+1\] So, \[{{a}^{2}}+1=({{a}^{2}}+2)(1-{{e}^{2}})\] \[\Rightarrow \] \[{{a}^{2}}+1=({{a}^{2}}+2).\frac{5}{6}\] \[\Rightarrow \] \[6{{a}^{2}}+6=5{{a}^{2}}+10\] \[\Rightarrow \] \[{{a}^{2}}=10-6=4\] \[\Rightarrow \] \[a=\pm 2\] \[\therefore \] \[Latusrectum=\frac{2({{a}^{2}}+1)}{\sqrt{{{a}^{2}}+2}}=\frac{2\times 5}{\sqrt{6}}=\frac{10}{\sqrt{6}}\]You need to login to perform this action.
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