<s>[INST] If $PA - PB = \text{constant}$ then the locus of $P$ will be a hyperbola. The sound of a cannon firing is heard one
second later at a position B than at position A.
If the speed of sound is uniform, then 
(A) The positions A and B are foci of a
hyperbola, with cannon's position on one
branch of the hyperbola
(B) the position A and B are foci of an ellipse
with cannon's position on the ellipse
(C) One of the positions A,B is focus of a
parabola with cannon's position on the
parabola
(D) It is not possible to describe the positions
of A, B and the cannon with the given
information 
My Attempt: If I suppose the position of A and B are respectively $(c,0)$ and $(0,-c)$ and $2a = 343\ $m, then we will get the locus of $P$ as follows $$\frac{x^2}{a^2} - \frac{y^2}{(c^2 - a^2)} = 1~.$$
If $c^2 > a^2$ we are done but if not then what will happen? 
Explicitly saying if the speed of the sound is greater than the distance between two points then what will happen?
Can anyone please help me  if I have gone wrong anywhere?
 [/INST] EDIT This was my first answer to the original question. OP change his question afterward, so I'll update my answer.
The geometric definition of an hyperbola is : 

the points such that the difference of the distance to two foci is
  constant. This difference is equal to the distance between the apex of the hyperbola

If A and B are the foci, the cannon is somewhere on an hyperbola.

The geometric definition I gave is equivalent to the following algebric definition of the horizontal hyperbola

If the foci of the hyperbola are located at $(c, 0)$ and $(-c, 0)$, then the hyperbola is given by the equation $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}=1 \qquad \text{where }c^2 = a^2+b^2$$

Proof
Starting with the geometric definition.  Let $A(c, 0)$ and $B(-c,0)$, the foci, and $P(x, y)$ a point on the hyperbola. Then, the difference of the distances is equal to $2a$.
$$\vert d(P, A) - d(P, B)\vert = 2a$$
$$\left\vert \sqrt{(x-c)^2+(y-0)^2} - \sqrt{(x-(-c))^2+(y-0)^2}\right\vert = 2a$$
If we square each side, it become
$$(x-c)^2+y^2 - 2\sqrt{(x-c)^2+y^2}\sqrt{(x+c)^2+y^2} + (x+c)^2+y^2 = 4a^2$$
Developping the $(x\pm c)^2$ and moving things around give us
$$x^2+y^2+c^2-2a^2=\sqrt{(x-c)^2+y^2}\sqrt{(x+c)^2+y^2}$$
We square once again to get rid of square roots.
$$(x^2+y^2+c^2-2a^2)^2=((x-c)^2+y^2)((x+c)^2+y^2)$$
Developping both side
$$x^4+y^4+c^4+4a^4 + 2x^2y^2+2x^2c^2-4a^2x^2+2y^2c^2-4a^2y^2-4a^2c^2 = x^4-2x^2c^2+c^4+y^4+2y^2x^2+2y^2c^2$$
Cancelling terms on each side and reorganise remaining terms
$$4a^4+4x^2c^2-4a^2x^2-4a^2y^2-4a^2c^2 = 0$$
$$x^2(c^2-a^2)-a^2y^2=a^2(c^2 - a^2)$$
Since $c^2 = a^2+b^2$
$$x^2b^2-a^2y^2=a^2b^2$$
Dividing everywhere by $a^2b^2$
$$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$
A similar proof could be done for a vertical hyperbola $\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$
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