Question

Let the locus of foot of the perpendicular drawn from $(1,-2)$ to the family of lines $(3+\lambda )x+(3\lambda -2)y=8-\lambda$ is $s=0$. If the line's joining origin to the point of intersections of $s=0$ with the line $x+by=1$ makes an angle of ${60}^{\circ }$ then sum of all possible values of $b$ is

• A. $\frac{3}{2}$
• B. $\frac{-3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{-5}{2}$

Answer 1

The correct option is B $\frac{-3}{2}$

Point of intersection of family of lines $(3+\lambda )x+(3\lambda -2)y=8-\lambda$ is $(2,-1)$
Let $p(x,y)$ be the required locus

$\therefore \left(m_{AP}\right)\left(m_{PB}\right)=-1$
$\Rightarrow (y+2)(y+1)+(x-1)(x-2)=0$
$\Rightarrow x^2+y^2-3x+3y+4=0$
Homogenising curve obtained with the line $x+by=1,$ we have
$x^2+y^2-3x(x+by)+3y(x+by)+4(x+by{)}^2=0$
$\Rightarrow 2x^2+(5b+3)xy+(4b^2+3b+1)y^2=0$
Angle between these lines is ${60}^{\circ }$
$\therefore \tan \theta =2\frac{\sqrt{h^2-ab}}{a+b}$
$\Rightarrow 3(a+b{)}^2=4(h^2-ab)$
$\Rightarrow 3(4b^2+3b+3{)}^2=(-7b^2+6b+1)$
$\Rightarrow 48b^4+72b^3+\cdots =0$
$\therefore$ sum of required values of $b=-\frac{\left(24\right)\left(3\right)}{48}=\frac{-3}{2}$