Question

Prove that the equation to the bisectors of the angle between the straight lines $a{x}^2+2hxy+b{y}^2=0$ is $h(x^2-y^2)+(b-a)xy=(a{x}^2-b{y}^2)\cos \omega$, the axes being inclined at an angle $\omega$.

Answer 1

Let $X^{{}^{\prime }}OX$ and $YO{Y}^{{}^{\prime }}$ be the axes inclined at an angle $\omega$. CD and AB be two straight lines represented by given equation $a{x}^2+2hxy+b{y}^2=0$ inclined at angles ${\theta }_1$ &${\theta }_2$ to x-axis respectively.
Let PQ and RS be the bisectors of the angles between AB and CD. If PQ is inclined at an angle $\theta$ to x axis, RS will be inclined at $(\frac{\pi }{2}+\theta )$
Again $\theta ={\theta }_1+\frac{1}{2}({\theta }_1-{\theta }_2)=\frac{{\theta }_1+{\theta }_2}{2}$
or $2\theta ={\theta }_1+{\theta }_2.....1$
Let CD be represented by $y=m_{1}x$ and AB by $y=m_{2}x$ Hence the combined equation will be $(y-m_{1}x)(y-m_{2}x)=0$
Comparing with given equation we get,
$m_1+m_1=-\frac{2h}{b}$ and $m_1{m}_2=\frac{a}{b}$
Now, $\tan 2\theta =\tan ({\theta }_1+{\theta }_2)$
$=\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}$
$=\frac{\frac{m_1\sin \omega }{1+m_1\cos \omega }+\frac{m_2\sin \omega }{1+m_2\cos \omega }}{1-\frac{m_1\sin \omega }{1+m_1\cos \omega }\frac{m_2\sin \omega }{1+m_2\cos \omega }}$
$\because$ If the line $y=m_{1}x$ makes an angle ${\theta }_1$ from x axis , then
$\tan {\theta }_1=\frac{m_1\sin \omega }{1+m_1\cos \omega }$
$=\frac{2m_1{m}_2\sin \omega \cos \omega +\{m_1+m_2\}\sin \omega }{1+\{m_1+m_2\}\cos \omega -m_1{m}_2({\sin }^2\omega -{\cos }^2\omega )}$
$\therefore \tan 2\theta =\frac{2a\sin \omega \cos \omega -2h\sin \omega }{b-2h\cos \omega -a({\sin }^2\omega -{\cos }^2\omega )}$
$\because m_1+m_2=-\frac{2h}{b},m_1{m}_2=\frac{a}{b}....2$
Again let PQ be $y=mx.....3$
As PQ is inclined at angle of $\theta$ to x-axis,$\tan \theta =\frac{m\sin \omega }{1+m\cos \omega }$
$\therefore \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$
$=\frac{2y\sin \omega (x+y\cos \omega )}{{(x+y\cos \omega )}^2-{\left(y\sin \omega \right)}^2}....4$
Equating RHS of $2$ and $4$ as LHS are same, we get
$\frac{2a\sin \omega \cos \omega -2h\sin \omega }{b-2h\cos \omega -a({\sin }^2\omega -{\cos }^2\omega )}=\frac{2y\sin \omega (x+y\cos \omega )}{{(x+y\cos \omega )}^2-{\left(y\sin \omega \right)}^2}$
On simplifying we get;-
$h(x^2-y^2)-(b-a)xy=(a{x}^2-b{y}^2)\cos \omega$