Question

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

• A. ${\sec }^{2}2\alpha =h$
• 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: (B), (D)

The correct options are
B $\cos \alpha =\sqrt{\frac{1+h}{2h}}$
D $\cot \alpha =\sqrt{\frac{h+1}{h-1}}$

Let equation of the lines given by

$x^2+2hxy+y^2=0$ be $y=m_{1}xandy=m_{2}x$. Since these

make an angle $\alpha$ with $y+x=0$ whose slope is -1.

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

$\Rightarrow m_1=\frac{\tan \alpha -1}{\tan \alpha +1}and{m}_2=\frac{\tan \alpha +1}{\tan \alpha -1}$

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

$=\frac{2{\sec }^2\alpha \times {\cos }^2\alpha }{\cos 2\alpha }=-2\sec 2\alpha =-2h$

$\Rightarrow \sec 2\alpha =h$

$\Rightarrow \cos 2\alpha =\frac{1}{h}\Rightarrow 2{\cos }^2\alpha -1=\frac{1}{h}$

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