question_answer

An equilateral triangle\[SAB\]is inscribed in the parabola\[{{y}^{2}}=4ax\]having its focus at\[S\]. If chord\[AB\]lies towards the lefit of\[S\], then side length of this triangle is

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,  \[AB=4a{{t}_{1}}=4a(2-\sqrt{3})\]