Question

Prove that the equation
$y^2(\cos \alpha +\sqrt{3}\sin \alpha )\cos \alpha -xy(\sin 2\alpha -\sqrt{3}\cos 2\alpha )+x^2(\sin \alpha -\sqrt{3}\cos \alpha )\sin \alpha -0$
represents two straight lines inclined at ${60}^o$ to each other.
Prove also that the area of the triangle formed with them by the straight line
$(\cos \alpha -\sqrt{3}\sin \alpha )y-(\sin \alpha +\sqrt{3}\cos \alpha )x+a=0$ is $\frac{a^2}{4\sqrt{3}},$
and that this triangle is equilateral.

Answer 1

$y^2(\cos \alpha =\sqrt{3}\sin \alpha )\cos \alpha -xy(\sin 2\alpha -\sqrt{3}\cos 2\alpha )+x^2(\sin \alpha -\sqrt{3}\cos \alpha )\sin \alpha =0$
So the given equation is a homogenous equation in second degree, it represents $2$ straight lines passing through origin
So $-(\sin 2\alpha -\sqrt{3}\cos 2\alpha )=(\sqrt{3}{\cos }^2\alpha -\sin \alpha )$
$=\sqrt{3}{\cos }^2\alpha -\sqrt{3}{\sin }^2\alpha -2\sin \alpha \cos \alpha$
$=(\sqrt{3}\cos \alpha +\sin \alpha )(\cos \alpha -\sqrt{3}\sin \alpha )$
The given equation can be factorized as
$(y\cos \alpha -x\sin \alpha )(y(\cos \alpha +\sqrt{3}\sin \alpha )-x(\sin \alpha -\sqrt{3}\cos \alpha ))=0$
Hence the given equation represents
$y\cos \alpha -x\sin \alpha =\mathrm{0....}\left(1\right)$
and $y(\cos \alpha +\sqrt{3}\sin \alpha )-x(\sin \alpha -\sqrt{3}\cos \alpha )=0$
If $\theta$ be the angle between them, then
$\tan \theta =\pm \frac{\frac{\sin \alpha -\sqrt{3}\cos \alpha }{\cos \alpha +\sqrt{3}\sin \alpha }-\frac{\sin \alpha }{\cos \alpha }}{1+\frac{\sin \alpha -\sqrt{3}\cos \alpha }{\cos \alpha +\sqrt{3}\sin \alpha }\frac{\sin \alpha }{\cos \alpha }}$
$=\pm \sqrt{3}$ (on simplification)
$\therefore \theta =60°$
The third given straight line is
$(\cos \alpha -\sqrt{3}\sin \alpha )y-(\sin \alpha +\sqrt{3}\cos \alpha )x+a=\mathrm{0.....}\left(3\right)$
The point of interaction of $\left(1\right)$ & $\left(2\right)$
Say $\left(A\right)$, is clearly $=(0,0)$
The point of interaction of $\left(1\right)$ & $\left(3\right)$
Say B will be
$=(\frac{a\cos \alpha }{\sqrt{3}},\frac{a\sin \alpha }{\sqrt{3}})$
If $\Phi$ be the angle between $\left(1\right)$ & $\left(3\right)$ then
$\tan \Phi =\pm \left\{\frac{\frac{\sin \alpha }{\cos \alpha }-\frac{\sin \alpha +\sqrt{3}\cos \alpha }{\cos \alpha -\sqrt{3}\sin \alpha }}{1+\frac{\sin \alpha }{\cos \alpha }\frac{\sin \alpha +\sqrt{3}\cos \alpha }{\cos \alpha -\sqrt{3}\sin \alpha }}\right\}=\pm \sqrt{3}$
$\therefore \Phi =60°$
So $2$ angles of $△ABC$ are $60°$ each ; hence $3^{rd}$ will also be $60°$
$\therefore$ The triangle is equilateral triangle
Area of an equilateral triangle
$=\frac{x^2\sqrt{3}}{9}$
Side AB
$x=\sqrt{\{{\left(\frac{a\cos \alpha }{\sqrt{3}}\right)}^2+{\left(\frac{a\sin \alpha }{\sqrt{3}}\right)}^2\}}$
$=\frac{a}{\sqrt{3}}$
The area of the given triangle is
$={\left(\frac{a}{\sqrt{3}}\right)}^2\frac{\sqrt{3}}{4}=\frac{a^2}{4\sqrt{3}}$