Question

If $x^2+2hxy+y^2=0$ represents the equation of the straight lines through the origin which make an angle $\alpha$ with the straight line $y+x=0$, then

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

Answer 1

Answer: (A), (B), (D)

$\sec 2\alpha =h$
$\cos \alpha =\sqrt{\frac{1+h}{2h}}$
$\cot \alpha =\sqrt{\frac{h+1}{h-1}}$

Equation of the lines given by $x^2+2hxy+y^2=0$ be $y=m_{1}x$ and $y=m_{2}x$.
$\Rightarrow m_1+m_2=-2h,m_1{m}_2=1$
Since these make an angle $\alpha$ with $y+x=0$ whose slope is $-1,$ so
$\frac{m_1+1}{1-m_1}=\tan \alpha ....\left(1\right)$
$\tan \alpha =\frac{-1-m_2}{1-m_2}....\left(2\right)$
.
From $\left(1\right)$ , $\frac{\tan \alpha +1}{\tan \alpha -1}=\frac{2}{2m_1}=\frac{1}{m_1}....\left(3\right)$
From $\left(2\right)$, $\frac{\tan \alpha -1}{\tan \alpha +1}=\frac{-2}{-2m_2}=\frac{1}{m_2}....\left(4\right)$
Adding $\left(3\right)$ and $\left(4\right)$, $\frac{m_1+m_2}{m_1{m}_2}=\frac{{(\tan \alpha +1)}^2+{(\tan \alpha -1)}^2}{{\tan }^2\alpha -1}$
$\Rightarrow \frac{-2h}{1}=\frac{2(1+{\tan }^2\alpha )}{-(1-{\tan }^2\alpha )}\Rightarrow 2h=2\sec 2\alpha$
$\Rightarrow \sec 2\alpha =h$
$\Rightarrow \cos 2\alpha =\frac{1}{h}\Rightarrow 2{\cos }^2\alpha -1=\frac{1}{h}$
$\Rightarrow \cos \alpha =\pm \sqrt{\frac{1+h}{2h}}$ and $\cot \alpha =\pm \sqrt{\frac{h+1}{h-1}}$