Question

# An equilateral triangle SAB is inscribed in the parabola ${y^2} = 4ax$ having its focus at S. If chord AB lies towards the left of S, then side length of the triangle isA. $2a\left( {2 - \sqrt 3 } \right)$B. $4a\left( {2 - \sqrt 3 } \right)$C. $a\left( {2 - \sqrt 3 } \right)$D. $8a\left( {2 - \sqrt 3 } \right)$

Hint: To find the length of the triangle, begin by considering two points, using these two points find out the slope of AS using the formula of slope and find the value of ${t_1}$.

Let $A\left( {a{t_1}^2,2a{t_1}} \right),B\left( {a{t_1}^2, - 2a{t_1}} \right)$

It is given that $\angle ASO = \dfrac{\pi }{6}$, therefore the slope of AS will be,
${m_{AS}} = \tan \left( {\dfrac{{5\pi }}{6}} \right)$

On using the formula to find slope which is, $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$, we get,
$\Rightarrow \dfrac{{2a{t_1}}}{{a{t_1}^2 - a}} = - \dfrac{1}{{\sqrt 3 }}$
$\Rightarrow {t_1}^2 + 2\sqrt 3 {t_1} - 1 = 0$
$\Rightarrow {t_1}^2 = - \sqrt 3 \pm 2$

Clearly, ${t_1} = - \sqrt 3 - 2$ is rejected.

Thus, ${t_1} = \left( {2 - \sqrt 3 } \right).$

Hence, $AB = 4a{t_1} = 4a\left( {2 - \sqrt 3 } \right)$

Option B is the correct answer.

Note: We started by assuming two points and then calculated the value of ${t_1}$using the formula of slope of AS. In this question, we rejected the negative value of ${t_1}$as it would not have matched with the given options.