Showing locus of points is a hyperbola

Question

I've having trouble understanding what the question is trying to ask. And I am not sure how to start to answer the question.

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The diagram below shows what happens for waves on the surface of a pond. If you drop a stone in the point at the point $F_1$ at time $T_1$, then a ripple will radiate outwards with constant speed $c$. At time $t$, this appears as a circle with centre $F_1$ and radius $c(t-T_1)$. If you drop another stone in the pond at the point $F_2$ at time $T_2$, then at time $t$ you will see another circle with centre $F_2$ and radius $c(t-T_2)$.

Show that the locus of points $P(t)$ where the waves at time $t$ meet is a single component of a hyperbola with foci $F_1$ and $F_2$ and length $|PF_1|-|PF_2|=c(T_2-T_1).$

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I was thinking maybe to show it algebraically but the course requires me to do it geometrically. I also know that I can prove it by showing $|PF_1|-|PF_2|$ to be some constant. But I'm not sure how to use it in the context of the question.

Answer 1

In the picture, the various colors indicate different times. There are only two wavefront circles at each given time.
At time $1$, that's the black dot at $F_2$ and the small black circle around $F_1$. They don't meet, so they don't contribute any points to the locus.
At time $2$, that's the two red circles, and the two red points they meet at will be part of the locus.
At time $3$, that's the two blue circles, and the two blue points they meet at will be part of the locus.
Time $4$ gets us the two green circles. Time $5$ gets us a large black circle around $F_2$, with the counterpart for $F_1$ not drawn in.

So, how can we determine if some $P$ in the plane is on the locus? The wave from $F_1$ reaches $P$ at time $t$ if the distance $PF_1$ is equal to $c(t-T_1)$ - the speed of the wave times the time since the wave started. The wave from $F_2$ reaches $P$ at time $t$ if the distance $PF_2$ is equal to $c(t-T_2)$. We want to know if these are both true for the same $t$, so we set up the system and solve it: \begin{align*}PF_1 &= c(t-T_1)\\ PF_2 &= c(t-t_2)\\ PF_1-PF_2 &= c(t-T_1-t+T_2) = c(T_2-T_1)\end{align*} Now, $c(T_2-T_1)$ is just a constant. The signed difference between the distances from our two points $F_1$ and $F_2$ is a constant, and that's the standard description of a component of a hyperbola with those two points as foci. Done.

(We would get the other component by reversing the sign of that constant $c(T_2-T_1)$)

Answer 2

Analytically, let the circles have equations

$$\begin{cases}(x+1)^2+y^2=(t+1)^2,\\(x-1)^2+y^2=t^2\end{cases}$$ and eliminate $t$ to get the trajectory of the intersection.

By subtraction,

$$4x=2t+1$$

and

$$(x-1)^2+y^2=\left(\frac{4x-1}2\right)^2$$

or

$$y^2-3x^2+\frac34=0$$ which describes an hyperbola.

Answer 3

1)$ F_1,F_2$, fixed points.

Locus of a point $P$ s.t.

$|PF_1|-|PF_2| = const. >0$,

is a hyperbola.

In your case:

$|P(t)F_1| -|P(t)F_2|= c(T_2-T_1) =const >0$.

2) Construct points on the hyperbola, given the foci $F_1,F_2$, and $ |PF_1|-|PF_2|= d>0$, constant .

Consider points $A(=F_1)$, $B (=F_2)$,

$A$ and $B$ are fixed, and a point $C(=P)$ to, be constructed s.t. $b-a = d$ fixed ,

i.e. :

Construct point $C$ of a $\triangle ABC$ with

base $AB$ , and $b-a =d$ given.

1)Draw a circle with radius $d$ around $A$. Pick any point $D$ on the circle.

2) Join $D$ and $B$.

3) Construct the perpendicular bisector of $DB$.

4) Extend $AD$ to intersect the bisector at $C$.

5) $\triangle ABC$ , has sides $AB$(basis), $a$, and $b$.

Reasoning:

6) $DC=BC = b$(by construction).

7) $AC= AD +DC=$

$(b-a)+a=a$.

Answer 4

It's well known that the sum of the distances of any point on an ellipse to the foci is a constant, the major axis (Apollonius, Conics III, 52). (Recall the method, mentioned by Descartes, of drawing an ellipse using two nails, some string, and a pencil.)

Correspondingly, on a hyperbola the difference of the distances of any point to the foci is constant, again the major axis (Conics, III, 51).

Since there is a time lag between when the two stones were dropped in the water, but the waves are presumed to be traveling at the same uniform speed, the difference in the distances of each wave from its starting point will be constant.

So if the converse of Conics III, 51 is true, the locus of wave intersections will be a hyperbola.