Question

The angle between the pair of lines $(x^2+y^2){\sin }^2\alpha ={(x\cos \theta -y\sin \theta )}^2$ is :

• A. $\theta$
• B. $2\theta$
• C. $\alpha$
• D. $2\alpha$

Answer 1

Answer: (D)

Given pair of lines of $(x^2+y^2)si{n}^2\alpha ={(x\cos \theta -y\sin \theta )}^2$
$\Rightarrow$ $x^2{\sin }^2\alpha +y^2{\sin }^2\alpha =x^2{\cos }^2\theta$
$+y^2{\sin }^2\theta -2xy\sin \theta \cos \theta$
$\Rightarrow$ $x^2({\sin }^2\alpha -{\cos }^2\theta )+y^2({\sin }^2\alpha -{\sin }^2\theta )$
$+2xy\sin \theta \cos \theta =0$
$\Rightarrow$ $x^2({\sin }^2\alpha -{\cos }^2\theta )+y^2({\sin }^2\alpha -{\sin }^2\theta )$
$+2(\sin \theta \cos \theta )xy=0$
On comparing with $a{x}^2+b{y}^2+2hxy=0,$
We get, $a={\sin }^2\alpha -{\cos }^2\theta ,b={\sin }^2\alpha -{\sin }^2\theta$
and $h=\sin \theta \cos \theta$
Let $\theta$ be the angle between the pair of lines. $\therefore$
$\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$
$=\left|\frac{2\sqrt{\begin{array}{rl}& {\sin }^2\theta {\cos }^2\theta -({\sin }^2\alpha -{\cos }^2\theta )\\ & \times ({\sin }^2\alpha -{\sin }^2\theta )\end{array}}}{{\sin }^2\alpha -{\cos }^2\theta +{\sin }^2\alpha -{\sin }^2\theta )}\right|$
$=\left|\frac{2\sqrt{\begin{array}{rl}& {\sin }^2\theta {\cos }^2\theta -{({\sin }^2\alpha )}^2+{\sin }^2\alpha \\ & {\sin }^2\theta +{\sin }^2\alpha {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta \end{array}}}{-(-1-2{\sin }^2\alpha )}\right|$
$=\left|\frac{2\sqrt{{\sin }^2\alpha ({\sin }^2\theta +{\cos }^2\theta )-{(si{n}^2\alpha )}^2}}{-\cos 2\alpha }\right|$
$=\left|\frac{2\sqrt{{\sin }^2\alpha (1-{\sin }^2\alpha )}}{-\cos 2\alpha }\right|$
$\Rightarrow$ $\tan \theta =\left|\frac{\sin 2\alpha }{\cos 2\alpha }\right|=\tan 2\alpha$
$\Rightarrow$ $\theta =2\alpha$