Question

$L:x^2+2gxy+y^2=0$ represents the equation of pair of straight lines passing through the origin. If $L$ makes an acute angle of $\theta$ with the straight line $y=x$, then

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

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\sec 2\theta =-g$
B $\cos \theta =\sqrt{\frac{g-1}{2g}}$
C $\tan \theta =\sqrt{\frac{g+1}{g-1}}$

$L:x^2+2gxy+y^2=0$
Let the equation of pair of lines be $y=m_{1}x$ and $y=m_{2}x$

$x^2+2gxy+y^2=0$
$\Rightarrow y=\frac{-2gx\pm \sqrt{4g^2{x}^2-4x^2}}{2}$
$\Rightarrow y=(-g\pm \sqrt{g^2-1})x$

$\text{Let}{m}_1=-g+\sqrt{g^2-1}\text{and}{m}_2=-g-\sqrt{g^2-1}$
Then $m_1+m_2=-2g$

$L$ makes an angle of $\theta$ with the straight line $y=x$
$\Rightarrow \frac{m_1-1}{1+m_1}=\tan \theta =\frac{1-m_2}{1+m_2}\Rightarrow \frac{1+\tan \theta }{1-\tan \theta }=m_1\text{and}\Rightarrow \frac{1-\tan \theta }{1+\tan \theta }=m_2$
$\Rightarrow m_1+m_2=\frac{(1+\tan \theta {)}^2+(1-\tan \theta {)}^2}{1-{\tan }^2\theta }$

$\Rightarrow 2\sec 2\theta =m_1+m_2=-2g\Rightarrow \sec 2\theta =-g\cos 2\theta =\frac{-1}{g}\tan 2\theta =\sqrt{g^2-1}\cos \theta =\sqrt{\frac{g-1}{2g}}\tan \theta =\sqrt{\frac{g+1}{g-1}}$