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 the value of $h$ is

• A. $\cos 2\alpha$
• B. $\sin 2\alpha$
• C. $\sec 2\alpha$
• D. $\csc 2\alpha$

Answer 1

Answer: (C)

$\sec 2\alpha$

Let equation of the lines given by $x^2+2hxy+y^2=0$ be $y=m_{1}x$ and $y=m_{2}x$.

Since these lines make an angel $a$ with $y+x=0$ whose slope is $-1$.

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

$\Rightarrow m_1=\frac{\tan a-1}{\tan a+1}$ and $m_2=\frac{\tan a+1}{\tan a-1}$

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

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