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 $\dfrac{\alpha}2$ in the clockwise direction. If in the new position, the slope of the line is $2-\sqrt3$ and its distance from the origin is $\dfrac1{\sqrt2}$, then the value of $3a^2	an^2\alpha - 2\sqrt3$ is:

Accepted Answer

$$\begin{aligned}
& \mathrm{m}_{\mathrm{PR}}=2-\sqrt{3}=\tan 15^{\circ} \
& \therefore \frac{\alpha}{2}=15^{\circ}
\end{aligned}$$

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

equation of PR:

$$\begin{aligned}
& y=\tan 15^{\circ}(x-a) \
& y=(2-\sqrt{3})(x-a)
\end{aligned}$$

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

$$\begin{aligned}
& \left|\frac{\sqrt{3} a-2 a}{\sqrt{4+3-4 \sqrt{3}+1}}\right|=\frac{1}{\sqrt{2}} \
& \frac{|a|(2-\sqrt{3})}{2 \sqrt{(2-\sqrt{3})}}=\frac{1}{\sqrt{2}} \
& |a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}=\sqrt{2}(\sqrt{2+\sqrt{3}}) \
& a^2=2(2+\sqrt{3}) \
& 3 a^2 \tan ^2 \alpha-2 \sqrt{3} \
& 3 \times(4+2 \sqrt{3}) \cdot \frac{1}{3}-2 \sqrt{3}=4
\end{aligned}$$

Solutions · 1