Question

Consider a general equation of degree $2$, as $\lambda x^2-10xy+12y^2+5x-16y-3=0$ For the value of $\lambda$ obtained for the given equation to be a pair of straight lines, if $\theta$ is the acute angle between $L_1=0$ and $L_2=0$ then $\theta$ lies in the interval

• A. $({45}^{\circ },{60}^{\circ })$
• B. $({30}^{\circ },{45}^{\circ })$
• C. $({15}^{\circ },{30}^{\circ })$
• D. $(0^{\circ },{15}^{\circ })$

Answer 1

The correct option is D $(0^{\circ },{15}^{\circ })$

Given pair of line

$\lambda x^2-10xy+12y^2+5x-16y-3=0$

on comparing above eq with general form of pair of eq we get

$a=\lambda ,h-5,b=12,g=\frac{5}{2},f=-8,c=-3$

Given eq is pair of eq so

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

$\lambda \times 12\times (-3)+2(-8)\left(\frac{5}{2}\right)(-5)-\lambda \times 64-12\left(\frac{25}{4}\right)-(-3)\left(25\right)=0$

$-36\lambda +200-64\lambda -75+75=0$

$-100\lambda +200=0$

$\lambda =2$

eq of pair becomes

$2x^2-10xy+12y^2+5x-16y-3=0$

$2x^2-(10y-5)x+(12y^2-16y-3)=0$

$x=\frac{10y-5\pm \sqrt{(10y-5{)}^2-8(12y^2-16y-3)}}{4}$

$4x=10y-5\pm \sqrt{100y^2+25-100y-96y^2+128y+24)}$

$4x=10y-5\pm \sqrt{4y^2+28y+49)}$

$4x=10y-5\pm \sqrt{(2y+7{)}^2}$

$4x=10y-5\pm (2y+7)$

$4x=10y-5+2y+7$ or $4x=10y-5-(2y+7)$

$4x-12y-2=0$ or $4x-8y+12=0$

$2x-6y-1=0$ or $2x-4y+6=0$

$L_1:2x-6y-1=0$

$L_2:2x-8y-6=0$

Slope of line $L_1,L_2$ $m_1=\frac{1}{3}$ and $m_2=\frac{1}{4}$

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

$\tan \theta =\left|\frac{\frac{1}{3}-\frac{1}{4}}{1+\frac{1}{3}\frac{1}{4}}\right|$

$\tan \theta =\frac{1}{13}$

$\therefore \theta ϵ(0^0,{15}^0)$