A) \[2a(2-\sqrt{3})\]
B) \[4a(2-\sqrt{3})\]
C) \[a(2-\sqrt{3})\]
D) \[8a(2-\sqrt{3})\]
Correct Answer: B
Solution :
Let\[A(at_{1}^{2},\,\,2a{{t}_{2}}),\,\,B(at_{2}^{2},\,\,-2a{{t}_{2}})\] We have, \[{{m}_{AS}}=\tan \left( \frac{5\pi }{6} \right)\] \[\Rightarrow \] \[\frac{2a{{t}_{1}}}{at_{1}^{2}-a}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \]\[t_{1}^{2}+2\sqrt{3}{{t}_{1}}-1=0\] \[\Rightarrow {{t}_{1}}=-\sqrt{3}\pm 2\] Clearly, \[{{t}_{1}}=\sqrt{3}-2\]is rejected. thus, \[{{t}_{1}}=(2-\sqrt{3})\], Hence, \[AB=4a{{t}_{1}}=4a(2-\sqrt{3})\]You need to login to perform this action.
You will be redirected in
3 sec