A line passing through the point P(a, 0) makes an acute angle with the positive x-axis.

Question

A line passing through the point P($a$, 0) makes an acute angle $\alpha $ with the positive x-axis. Let this line be rotated about the point P through an angle $\frac{\alpha }{2}$ in the clockwise direction. If in the new position, the slope of the line is $2-\sqrt{3}$ and its distance from the origin is $\frac{1}{\sqrt{2}}$, then the value of $3{a}^{2}{	an }^{2}\alpha -2\sqrt{3}$ is :

Accepted Answer

$\begin{gathered} {m}_{PR}=2-\sqrt{3}=	an {15}^{\circ } \ 	herefore \frac{\alpha }{2}={15}^{\circ } \end{gathered}$

$\Rightarrow \alpha ={30}^{\circ }$

equation of PR:

$\begin{gathered} y=	an {15}^{\circ }(x-a) \ y=(2-\sqrt{3})(x-a) \end{gathered}$

$\perp $ distance from origin $=\frac{1}{\sqrt{2}}$

$\begin{gathered} |\frac{\sqrt{3}a-2a}{\sqrt{4+3-4\sqrt{3}+1}}|=\frac{1}{\sqrt{2}} \ \frac{|a|(2-\sqrt{3})}{2\sqrt{(2-\sqrt{3})}}=\frac{1}{\sqrt{2}} \ |a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}=\sqrt{2}(\sqrt{2+\sqrt{3}}) \ {a}^{2}=2(2+\sqrt{3}) \ 3{a}^{2}{	an }^{2}\alpha -2\sqrt{3} \ 3	imes (4+2\sqrt{3})\cdot \frac{1}{3}-2\sqrt{3}=4 \end{gathered}$

Solutions · 1