Question

If the straight line $ax+by+p=0$ and $x\cos \alpha +y\sin \alpha =p$ enclosed an angle of $\frac{\pi }{4}$ and the line $x\sin \alpha -y\cos \alpha =0$ meets them at the same point, then $a^2+b^2$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

$2$

$x\sin \alpha -y\cos \alpha =0$

=> $x=\frac{y\cos \alpha }{\sin \alpha }$

$x\cos \alpha +y\sin \alpha =p$

Put $x=\frac{y\cos \alpha }{\sin \alpha }$, we get

=> $\frac{y{\cos }^2\alpha }{\sin \alpha }+y\sin \alpha =p$

=>$y=p\sin \alpha$

Similiarly, $x=p\cos \alpha$

$ax+by+p=0$

Put $x=p\cos \alpha$ and $y=p\sin \alpha$

=> $ap\cos \alpha +bp\sin \alpha +p=0$

=> $a\cos \alpha +b\sin \alpha =-1$ ---- Equation 1

Slope of $ax+by+p=0$, $m_1=-\frac{a}{b}$

Slope of $x\cos \alpha +y\sin \alpha =p$, $m_2=-\frac{\cos \alpha }{\sin \alpha }$

Also $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

=> $\tan \frac{\pi }{4}=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

=> $|m_1-m_2|=|1+m_1{m}_2|$

=> $|\frac{a}{b}-\frac{\cos \alpha }{\sin \alpha }|=|1+\frac{a\cos \alpha }{b\sin \alpha }|$

=> $|a\sin \alpha -b\cos \alpha |=|b\sin \alpha +a\cos \alpha |$

=> $|a\sin \alpha -b\cos \alpha |=1$ ---- Equation 2

Squaring and adding equation 1 and equation 2, we get

$(a^2{\cos }^2\alpha +b^2{\sin }^2\alpha +2ab\sin \alpha \cos \alpha )+(b^2{\cos }^2\alpha +a^2{\sin }^2\alpha -2ab\sin \alpha \cos \alpha )=1+1$

=> $a^2+b^2=2$

Option $\left(C\right)$