Question

The angle between the straight lines $x\cos \alpha +y\sin \alpha =p$ and $ax+by+p=0$ is $\pi /4$. They meet the straight line $x\sin \alpha -y\cos \alpha =0$ in the same point; then prove that $a^2+b^2=2$.

Answer 1

Two of the lines are evidently perpendicular and the line $ax+by+p=0$ makes an angle of ${45}^o$ with one of them and hence it is bisector of the two lines.
The bisectors of given lines are
$\frac{x\cos \alpha +y\sin \alpha }{1}=\pm \frac{x\cos \alpha -y\sin \alpha }{1}$
or $x(\cos \alpha -\sin \alpha )+y(\sin \alpha +\cos \alpha )-p=0$
compare with $ax+by+p=0$
$\frac{\cos \alpha -\sin \alpha }{a}=\frac{\sin \alpha +\cos \alpha }{b}=-1$
$\therefore -a\cos \alpha -\sin \alpha ,-b\sin \alpha +\cos \alpha$
$\therefore a^2+b^2=2$