Question

If the angle $2\theta$ is acute, then the acute angle between $x^2(\cos \theta -\sin \theta )+2xy\cos \theta +y^2(\cos \theta +\sin \theta )=0$is

• A. $2\theta$
• B. $\frac{\theta }{3}$
• C. θ
• D. $\frac{\theta }{2}$

Answer 1

The correct option is C

θ

Explanation for Correct Option:

Step 1: Apply trigonometric identities

$x^2(\cos \theta -\sin \theta )+2xy\cos \theta +y^2(\cos \theta +\sin \theta )=0.....\left(\mathrm{i}\right)$

General equation: $a{x}^2+2hxy+b{y}^2=0$

Comparing equation (i) with the general equation

$a=\cos \theta –\sin \theta ,b=\cos \theta +\sin \theta$ and $h=\cos \theta$

Step 2:Let $\alpha$ be the acute angle between the lines. Then,

$$\begin{gathered} \Rightarrow \tan \alpha =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right| \\ \Rightarrow \tan \alpha =\left|\frac{2\sqrt{{\cos }^2\theta –\left({\cos }^2\theta +{\sin }^2\theta \right)}}{2\cos \theta }\right|\left[\because {\cos }^2\theta +{\sin }^2\theta =1\right] \\ \Rightarrow \tan \alpha =\left|\frac{2\sqrt{{\cos }^2\theta –1}}{2\cos \theta }\right|\left[\because {\cos }^2\theta –1={\sin }^2\theta \right] \\ \Rightarrow \tan \alpha =\left|\frac{2\sin \theta }{2\cos \theta }\right| \\ \Rightarrow \tan \alpha =\left|\mathrm{ta}n\theta \right| \\ \Rightarrow \alpha ={\tan }^{-1}\mathrm{ta}n\theta \\ \Rightarrow \alpha =\theta \end{gathered}$$

Hence option (3) is Correct