Question

The equation of one of the lines represented by the equation $x^2-2xycot\theta -y^2=0$, is

• A. $x–ycot\theta =0$
• B. $x+y\tan \theta =0$
• C. $y\sin \theta +x(\cos \theta +1)=0$
• D. $x\cos \theta +y(\sin \theta +1)=0$

Answer 1

The correct option is C

$y\sin \theta +x(\cos \theta +1)=0$

Explanation for the correct options:

Step 1: Finding the value of $m$

The given equation of line is $x^2-2xycot\theta -y^2=0$

Dividing both sides by $x^2$, we get

$\Rightarrow 1–2\left(\frac{y}{x}\right)cot\theta –{\left(\frac{y}{x}\right)}^2=0....\left(i\right)$

Let the equation of the line be,

$y=mx....\left(ii\right)$

$\Rightarrow m=\frac{y}{x}$

Therefore, from equation $\left(i\right)$

$$\begin{gathered} \Rightarrow 1–2mcot\theta –m^2=0 \\ \Rightarrow m^2+2mcot\theta –1=0 \end{gathered}$$

Step 2: Solving this equation by using the quadratic formula

$\begin{array}{rcl}m& =& -\frac{2cot\theta \pm \sqrt{(4co{t}^2\theta +4)}}{2}\\ & =& -cot\theta \pm \sqrt{(co{t}^2\theta +1)}\\ & =& -cot\theta \pm \cos ec\theta \\ m& =& \left(-cot\theta +\cos ec\theta \right)\text{or}\left(-cot\theta –\cos ec\theta \right)\end{array}$

Step 3: Finding the equation of line

Substituting the value of $m$ in equation $\left(ii\right)$

$\begin{array}{rcl}y& =& (-cot\theta +\cos ec\theta )x\text{and}y=(-cot\theta –\cos ec\theta )x\\ y& =& \left[\left(\frac{-\cos \theta }{\sin \theta }\right)+\left(\frac{1}{\sin \theta }\right)\right]x\text{and}y=\left[\left(\frac{-\cos \theta }{\sin \theta }\right)-\left(\frac{1}{\sin \theta }\right)\right]x\\ y& =& \left[\frac{(1–\cos \theta )}{\sin \theta }\right]x\text{and}y=\left[\frac{-(1+\cos \theta )}{\sin \theta }\right]x\\ y\sin \theta & =& (1–\cos \theta )x\text{and}y\sin \theta =-(1+\cos \theta )x\\ y\sin \theta –x(1–\cos \theta )& =& 0\text{and},\\ y\sin \theta +x(1+\cos \theta )& =& 0\end{array}$

Hence, option (C) is the correct answer.