Question

If the axes be inclined at an angle $\omega$, shew that the equation $x^2+2xy\cos \omega +y^2\cos 2\omega =0$ represents a pair of perpendicular straight lines.

Answer 1

Let the given equation
$x^2+2xy\cos \omega +y^2\cos 2\omega =\mathrm{0....1}$
represent two straight lines $y=m_{1}x$ and $y=m_{2}x$
Then the combined equation will be
$(y-m_{1}x)(y-m_{2}x)=\mathrm{0.....2}$
Comparing the co efficients of different terms in $1$ and $2$
$\frac{1}{m_1{m}_2}=\frac{2\cos \omega }{-(m_1+m_2)}=\frac{\cos 2\omega }{1}$
$\therefore m_1{m}_2\frac{1}{\cos 2\omega }$ and $m_1+m_2=-\frac{2\cos \omega }{\cos 2\omega }$
If the two lines given by $2$ are perpendicular, then $1+(m_1+m_2)\cos \omega +m_1{m}_2=0$
Putting the calue of $m_1+m_2$ & $m_1{m}_2$ in LHS, we have
$1-\frac{2\cos \omega }{\cos 2\omega }\cos \omega +\frac{1}{\cos 2\omega }=0$
$=\frac{1}{\cos 2\omega }(\cos 2\omega -2{\cos }^2\omega +1)=0$
$\because \cos 2\omega =2{\cos }^2\omega -1$
Hence, the lines represented by $1$ are perpendicular to each other.