Question

The equation of one of the line represented by the equation
$x^2+2xy\cot \theta -y^2=0$ is

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

Answer 1

Answer: (A)

$x-y\cot \theta =0$

$x^2+\left(2\cot \theta \right)xy-y^2=0$

There are pair of straight lines

Passing through $h(0,0)$

Let be lines be $y=m_{1}x$

$y=m_{2}x$

Pair of lines:-

$(y-m_{1}x)(y-m_{2}x)=0$

As $y^2-xy(m_1+m_2)+\left(m_1{m}_2\right)x^2=0$

Comparing with $y^2=xy\left(2\cot \theta \right)-x^2=0$

$m_1{m}_2=-1$

$m_1+m_2=2\cot \theta$

Solving,

$m_1=\cot \theta +cosec\theta$

$m_2=\cot \theta +cosec\theta$

$\Rightarrow m_1=\frac{1+\cos \theta }{\sin \theta },m_2=\frac{\cos \theta -1}{\sin \theta }$

$\Rightarrow linesR\left\{\begin{array}{c}y\left(\sin \theta \right)=(1+\cos \theta )x\\ y\left(\sin \theta \right)=(\cos \theta -1)x\end{array}\right\}$