find minimum value of PQ

Question

Consider the locus of the complex number $'z'$ in the argand plane is given by $Re(z)-2=|z-7+2i|$.Let P($z_1$) and Q($z_2$) be two complex numbers satisfying the given locus and also satisfying $arg\bigg(\dfrac{z_1-(2+i\alpha)}{z_2-(2+i\alpha)}\bigg)=\dfrac{\pi}{2}$($\alpha\in R$) then the minimum value of PQ(distance)

My try:

let $z=x+iy$

from the locus of $z$ given,we get $(x-2)=\bigg|(x-7)+i(y+2)\bigg|$

$\implies$ $(x-2)^2=(x-7)^2+(y+2)^2$

$\implies$$(y+2)^2=5(2x-9)$

moving on we have $2$nd equation, $arg\bigg(\dfrac{z_1-(2+i\alpha)}{z_2-(2+i\alpha)}\bigg)=\dfrac{\pi}{2}$

which means that the value in the argument lies on positive side of $y-axis$

$\dfrac{z_1-(2+i\alpha)}{z_2-(2+i\alpha)}=ib$ for $b>0,b\in R$ plugging in $z_1$=$x_1+iy_1$ and $z_2=x_2+iy_2$, I got,

$x_1-2=-b(y_2-\alpha)$

$b(x_2-2)=y_1-\alpha$

how to proceed further$?$Any better way to solve equation $2$ $?$

Answer 1

As we know , the locus is a parabola , with vertex at (9/2,-2) .Now if we consider (2,alpha) as A , then by the giver conditions , PA is perpendicular to QA . Also the directrix of the given parabola is X=2 . As we know that the locus of point of intersection perpendicular tangents is directrix , we know that A lies on the directrix , and as a chord passing through the end points of perpendicular tangents PASSES THROUGH THE FOCUS , PQ passes through the focus . Therefore , minimum value of focal chord PQ is the length of Latus rectum , that is 10 in our case , so the answer is 10.