A is a given point and P is any point on a given straight line. If AQ = AP and AQ makes a constant angle with AP, find the locus of Q.

Question

I've been thinking a lot about this question for a while now, I checked various books on how one can find the locus of something, but I just can't understand. This is not a "homework question" but a question that has been bugging me for a while now. Could anyone tell me what this question even means? It seems to be forming a strange curve which doesn't exactly resemble an arc of a circle.

Answer 1

Draw out the figure as follows :

Begin by shifting origin to the point A.

Now, let the equation of the given straight line by $xcos(a) + ysin(a) = p$,

(here a & p are fixed clearly)

i.e. perendicular distance from A to the line = P

Now take any point P on this line such that :

it makes some angle '$t$' with the shortest distance from A to $xcos(a) + ysin(a) = p$.

Thus, our point Q lies on circle with radius $p.sec(t)$.

Assume some fixed angle (in counter-clockwise sense) that AQ makes with AP.

Call it '$b$'.

$\implies Q $ = $((p.sec(t)cos(a+b+t)$) , $(p.sec(t)sin(a+b+t)))$ = $(x,y)$ ;

$\implies x^2 + y^2$ = $(p.sec(t))^2$ ----------------(1)

$\implies x = p(sin(a+b)sec(t) - cos(a+b))$ ;

$\implies \displaystyle\frac{x+ p.cos(a+b)}{p.sin(a+b)}$ = $sec(t)$ ----------(2)

So, using (1) & (2), we wet the desired locus of Q ,i.e.

($(x+h)^2 + (y+k)^2$)($sin(a+b))^2$ = $((x+h)$ + $pcos(a+b))^2$

Note that here : $A = (h,k)$.