Question

An equilateral triangle PQR is formed where $P(1,3)$ is fixed point and Q is moving point on line $x=3$.Find the locus of $R$.

Answer 1

From figure,

Slope of $PR$ is $(\theta \pm {60}^{\circ })$

Let the coordinates of $R$ be $(h,k)$

$\therefore \frac{h-1}{\cos \theta \le {60}^{\circ }}=\frac{k-2}{\sin \theta \le {60}^{\circ }}=2\sec \theta$

$h=1+2\cos (\theta \pm {60}^{\circ })\sec \theta$

$⟹h=1+2[\frac{1}{2}\pm \frac{\sqrt{3}}{2}\tan \theta ]$ -----(1)

$k=2+2\sin (\theta \pm {60}^{\circ })\sec \theta$

$⟹k=2+2[\frac{\tan \theta }{2}\pm \frac{\sqrt{3}}{2}]$ -----(2)

Eliminating $\theta$ from (1) and (2),

$\frac{h-2}{\sqrt{3}}=\pm \tan \theta$

$k-2\pm \sqrt{3}=\tan \theta$

$⟹\frac{h-2}{3}=\pm (k-2\pm \sqrt{3})$

Locus is $(x-2)=\pm \sqrt{3}(y-2\pm \sqrt{3})$