Question

If $x^2-2hxy+y^2=0$ represents the equation of pair of straight lines both of which make an angle $\theta$ with the straight lines $x+y=2,$ then

• A. $\tan \theta =\sqrt{\frac{h-1}{h+1}}$
• B. $\cos 2\theta =h$
• C. $\cot \theta =\sqrt{\frac{h-1}{h+1}}$
• D. $\cos \theta =\sqrt{\frac{h+1}{2h}}$

Answer 1

The correct option is C $\cot \theta =\sqrt{\frac{h-1}{h+1}}$

$y=m_{1}x$ and $y=m_{2}x$
from $x^2-2hxy+y^2=0$
$m_1+m_2=2h{m}_1{m}_2=1$

$\frac{m_1+1}{1-m_1}=\tan \theta =-\frac{m_2+1}{1-m_2}$

$\Rightarrow m_1=\frac{\tan \theta -1}{\tan \theta +1};m_2=\frac{\tan \theta +1}{\tan \theta -1}$

$m_1+m_2=2h=\frac{(\tan \theta -1{)}^2+(\tan \theta +1{)}^2}{{\tan }^2\theta -1}$

$\Rightarrow \frac{{\sec }^2\theta -2\tan \theta +{\sec }^2\theta +2\tan \theta }{{\tan }^2\theta -1}$

$2h=\frac{2}{-\cos 2\theta }$

$2h=-2\sec 2\theta$
$h=-\sec 2\theta$
$-\frac{1}{h}=\cos 2\theta =2{\cos }^2\theta -1$
$\cos \theta =\sqrt{\frac{h-1}{2h}}$

$\tan \theta =\sqrt{\frac{h+1}{h-1}}$

$\Rightarrow \cot \theta =\sqrt{\frac{h-1}{h+1}}$