Suppose that a hawk, whose initial position is $(a,0)=(6000,0)$ on the $x$-axis, spots a pigeon at $(0,-2000)$ on the $y$-axis. Suppose that the pigeon flies at a constant speed of 10 ft/sec in the direction of the $y$-axis (oblivious to the hawk), while the hawk flies at a constant speed of 20 ft/sec, always in the direction of the pigeon.

The problem is to find an equation for the flight path of the hawk (the curve of pursuit) and to find the time and place where the hawk will catch the pigeon. Assume that in this problem all distances are measured in feet and all times measured in seconds. Leave out all dimensions from your answers.

Consider the diagram above (click on it for a better view) which represents the situation at an arbitrary time $t$ during the pursuit. The points $P$ and $Q$ represent the positions of the hawk and pigeon respectively at that time instant $t$, with $y=f(x)$ representing the flight path of the hawk.

The pigeon's position $Q=(0,g(t))$ is given by the following function of time
$g(t)$ =

The fact that the hawk is always headed in the direction of the pigeon means that the line $PQ$ is tangent to the pursuit curve $y=f(x)$. This tells us that $\frac{dy}{dx}=h(x,y,t)$ where
$h(x,y,t)$ =
(Your answer must involve the three variables $x$, $y$, and $t$.)

If we solve the equation $$p = h(x,y,t),$$ where $p=\frac{dy}{dx}$, for time we obtain that $t=k(x,y,p)$ where
$k(x,y,p)=$ =
(Your answer must involve the three variables $x$, $y$, and $p$, where $p$ stands for $\frac{dy}{dx}$. )