Question

Angle between the lines represented by the equation $x^2+2xysec\theta +y^2=0,$is

• A. $\theta$
• B. $2\theta$
• C. $\frac{\theta }{2}$
• D. none of these

Answer 1

The correct option is A

$\theta$

Explanation for the correct option:

Step 1. Given that $x^2+2xysec\theta +y^2=0$

Comparing this equation with the general equation $a{x}^2+2hxy+b{y}^2=0$, We get

$a=1,$

$b=1$ and

$2h=2sec\theta$

$\Rightarrow$ $h=sec\theta$

We know the angle between the two lines given by

$\mathit{\alpha }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{2\sqrt{h^2-ab}}{a+b}\right)$

Step 2. Put the value of $h=sec\theta$, we get

$\begin{array}{rcl}tan\alpha & =& \frac{2\sqrt{h^2-ab}}{\left(a+b\right)}\\ & =& \frac{2\sqrt{se{c}^2\theta -\left(1\times 1\right)}}{\left(1+1\right)}\\ & =& \frac{2\sqrt{se{c}^2\theta -1}}{2}\\ & =& \sqrt{se{c}^2\theta -1}\\ & =& \sqrt{{\tan }^2\theta }\end{array}$

$\Rightarrow$$\begin{array}{rcl}\tan \alpha & =& \tan \theta \end{array}$

$\Rightarrow$ $\begin{array}{rcl}\alpha & =& {\tan }^{-1}\left(\tan \theta \right)\\ & =& \mathit{\theta }\end{array}$

Hence, option (A) is correct.