Question

If $y=mx$ bisects the angle between the lines $y=x^2({\sin }^2\theta +{\tan }^2\theta )+2xy\cos \theta +y^2{\sec }^2\theta =0$ when $\theta =\frac{\pi }{3}.$ If the value of $m$ is $\frac{a\pm \sqrt{b}}{4},$ then $b-10a=?$

Answer 1

Pair of straight lines:
$a{x}^2+2hxy+b{y}^2=0$
Equation of pair of straight lines which bisects the angle between these two lines:
$\frac{x^2-y^2}{a-b}=\frac{xy}{h}$
$a={\sin }^2\theta +{\tan }^2\theta$
$b={\sec }^2\theta$
$h=\cos \theta$
$\frac{x^2-y^2}{({\sin }^2\theta +{\tan }^2\theta )-{\sec }^2\theta }=\frac{xy}{\cos \theta }$
$\Rightarrow x^2-y^2=-\frac{xy}{2}$
$y=mx$ is one of the bisectors.
$x^2-m^2{x}^2=-\frac{m{x}^2}{2}$
$\Rightarrow 2m^2-m-2=0$
$m=\frac{1\pm \sqrt{17}}{4}$
$a=1,b=17\Rightarrow 17-10=7$