Question

Let the equation $x+\lambda y-2\sin \theta +\lambda (\cos \theta -1)=0$ represents a family of non-parallel lines for $a\theta$ $($where $\theta \in [0,\frac{\pi }{2}])$ Which passing through a fixed point (p, q) and distance of a point $(0,1)$ from the point $(p,q)$ is the greatest, then find the value of $(p+q)$.

Answer 1

$x-2sin\theta +\lambda (y+cos\theta -1)=0$

$⟹L_1$ is $x=2sin\theta$ and $L_2$ is $y=1-cos\theta$

Since the intersection point of $L_1$ and $L_2$ is the point of intersection of family of lines $L_1+\lambda L_2=0$,

$\therefore (p,q)⟹(2sin\theta ,1-cos\theta )$

The distance between $(p,q)$ and $(0,1)$ is

$d=\sqrt{(2sin\theta -0{)}^2+(1-cos\theta -1{)}^2}$

$d=\sqrt{4si{n}^2\theta +co{s}^2\theta }$

$d=\sqrt{3si{n}^2\theta +1}$

Now, as maximum value of $sin\theta$ occurs at $\theta =\frac{\pi }{2}$,

$d_{max}=2$ and $(p,q)⟹(2,1)$.

$⟹p+q=2+1=3$