Question

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

Answer 1

Given: $x^2+2hxy+y^2=0$
Since these lines make an angle $\alpha$ with $y+x=0$, so angle between the lines is $\pi -2a$

The angle between the lines is
$\tan (\pi -2a)=\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|\Rightarrow -\tan 2a=\left|\frac{2\sqrt{h^2-1}}{1+1}\right|\Rightarrow \tan 2a=\pm \sqrt{h^2-1}$

We know, $2\alpha \in (0,\pi )\Rightarrow \alpha \in (0,\frac{\pi }{2})$

$\Rightarrow \sec 2\alpha =\pm \sqrt{1+{\tan }^{2}2\alpha }\Rightarrow \sec 2\alpha =\pm \sqrt{1+h^2-1}\Rightarrow \sec 2\alpha =\pm h\Rightarrow \cos 2\alpha =\pm \frac{1}{h}\Rightarrow 2{\cos }^2\alpha -1=\pm \frac{1}{h}\Rightarrow \cos \alpha =\sqrt{\frac{h\pm 1}{2h}}\Rightarrow \sin \alpha =\sqrt{\frac{h\mp 1}{2h}}\Rightarrow \cot \alpha =\sqrt{\frac{h+1}{h-1}},\sqrt{\frac{h-1}{h+1}}$

Hence, options $\left(A\right),\left(B\right)$ and $\left(D\right)$ are correct.