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

The correct option is D

$2\alpha$

Explanation for correct option:

Finding the angle between the pair of lines:

Given, the equation of pair of lines

$(x^2+y^2){\sin }^2\alpha ={(x\cos \theta -y\sin \theta )}^2$

Expanding the above equation, we get

$$\begin{gathered} x^2{\sin }^2\alpha +y^2{\sin }^2\alpha =x^2{\cos }^2\theta -2xy\left(\sin \theta \right)\left(\cos \theta \right)+y^2{\sin }^2\theta \\ {x}^2{\sin }^2\alpha -x^2{\cos }^2\theta +y^2{\sin }^2\alpha -y^2{\sin }^2\theta +2xy\sin \theta \cos \theta =0 \\ \left({\sin }^2\alpha -{\cos }^2\theta \right)x^2+\left({\sin }^2\alpha -{\sin }^2\theta \right)y^2+2xy\sin \theta \cos \theta =0\to \left(i\right) \end{gathered}$$

The above equation is of the format $a{x}^2+b{y}^2+2hxy=0\to \left(ii\right)$ ,.

Comparing the equations $\left(i\right)\text{and}\left(ii\right)$, we get the values of

$a={\sin }^2\alpha -{\cos }^2\theta$

$b={\sin }^2\alpha -{\sin }^2\theta$

$h=\sin \theta \cos \theta$

Finding the angle between the pair of two lines using the formula $\tan \varphi =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$.

Therefore,

$\begin{array}{l}\tan \varphi =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|\\ =\left|\frac{2\sqrt{(\sin \theta \cos \theta {)}^2-\left({\sin }^2\alpha -{\cos }^2\theta \right)\left({\sin }^2\alpha -{\sin }^2\theta \right)}}{{\sin }^2\alpha -{\cos }^2\theta +{\sin }^2\alpha -{\sin }^2\theta }\right|\\ =\left|\frac{2\sqrt{{\sin }^2\theta {\cos }^2\theta -{\sin }^2\alpha {\sin }^2\alpha +{\cos }^2\theta {\sin }^2\alpha +{\sin }^2\alpha {\sin }^2\theta -{\cos }^2\theta {\sin }^2\theta }}{{\sin }^2\alpha -{\cos }^2\theta +{\sin }^2\alpha -{\sin }^2\theta }\right|\\ =\left|\frac{2\sqrt{-{\sin }^4\alpha +{\sin }^2\alpha \left({\cos }^2\theta +{\sin }^2\theta \right)}}{2{\sin }^2\alpha -\left({\cos }^2\theta +{\sin }^2\theta \right)}\right|\\ =\left|\frac{2\sqrt{-{\sin }^4\alpha +{\sin }^2\alpha \times 1}}{2{\sin }^2\alpha -\left(1\right)}\right|\left[\because {\cos }^2\theta +{\sin }^2\theta =1\right]\\ =\left|\frac{2\sqrt{{\sin }^2\alpha \left(1-{\sin }^2\alpha \right)}}{\cos 2\alpha }\right|\left[\because 2{\sin }^2\alpha -1=\cos 2\alpha \right]\\ =\left|\frac{2\sqrt{{\sin }^2\alpha \left({\cos }^2\alpha \right)}}{\cos 2\alpha }\right|\left[\because 1-{\sin }^2\alpha ={\cos }^2\alpha \right]\\ =\left|\frac{2\sin \alpha \cos \alpha }{\cos 2\alpha }\right|\\ =\left|\frac{\sin 2\alpha }{\cos 2\alpha }\right|[\because 2\sin \alpha \cos \alpha =\sin 2\alpha ]\\ =|\tan 2\alpha |\end{array}$

$$\begin{gathered} ⟹\tan \varphi =\tan 2\alpha \\ \varphi =2\alpha \end{gathered}$$

Therefore, the angle between the pair of lines is $2\alpha$

Hence, option (D) is the correct answer.