A graph $y=\sqrt{x}$ is plotted on a Cartesian plane. P is a point of the curve, except the origin and Q is the point where the perpendicular bisector of OP intersects the x-axis.
As P tends to O, does Q have a limiting position?
I have attempted to find a general expression for Q by letting P be (a,b) then finding the midpoint. But I'm unable to do so.
Please guide me on how I might solve this question. Thank you!
On
So $P = \left(x, \sqrt{x}\right)$, so the slope of $OP$ is $m=\sqrt{x}/x$ and the perpendicular $\ell$ will have slope $-1/m = -\sqrt{x}$ and pass through the middle of $OP$, which is $\left(x/2, \sqrt{x}/2\right)$.
It remains to find the $x$-intercept of $\ell$, can you do that?
On
Hint: Point $P$ will have coordinates $(x, \sqrt{x})$. Coordinates of the midpoint $M$ of $OP$ will be $(x/2, \sqrt{x}/2)$ Now you need to find the slope of the line $PO$ and then find an equation of a line perpendicular to $PO$ which goes thru point $M$. Then find out where perpendicular will intersect axis $x$. $P$ tends to $O$ means $x$ tends to zero.
Or you can also think when it's going to be impossible to drop a perpendicular to axis $x$ that goes thru $M$? Maybe when $PO$ is already perpendicular to axis $x$?
To avoid confusion, and to distinguish the point from the curve $y=\sqrt{x}$, let $P=(t,\sqrt{t})$.
Midpoint of $OP$, $M=(t/2, \sqrt{t}/2)$.
Slope of $OP$ is $\;\;\dfrac 1{\sqrt{t}}\;\;$ therefore slope of perpendicular bisector is $\;\;{-\sqrt{t}}\;$.
Equation of perpendicular bisector is $$y-\frac {\sqrt{t}}2=-\sqrt{t}\left(x-\frac t2\right)\\ y=-\sqrt{t}\left(x-\frac {t+1}2\right)$$
At $Q$, $y=0$, so $x=\frac {t+1}2$, i.e. $Q=\left(\dfrac {t+1}2,0\right)$.
As $P$ approaches $O$, $t\to 0$, hence $Q\to \color{red}{\left(\dfrac 12,0\right)}$
See desmos implementation here.