I have copied a solution I gave to a similar problem, which has the coordinate system defined such that the pursuer begins at the origin:
An interesting geometric model arises when one tries to determine the path of a pursuer chasing its "prey." This path is called a curve of pursuit. These problems were analyzed using methods of calculus circa 1760 (more than two centuries after Leonardo da Vinci had considered them).
The simplest problem is to find the curve along which a vessel moves when pursuing another vessel that flees along a straight line, assuming the speeds of the two vessels are constant.
Let's assume car $N$, traveling at speed $\beta$, is pursuing car $M$, which is traveling at speed $\alpha$. In addition, assume that car $N$ begins (at time $t=0$) at the origin and pursues car $M$, which begins at the point $(b,0)$ where $0<b$ and travels down the line $x=b$.
After $t$ hours, car $N$ is located at the point $P(x,y)$, and car $M$ is located at the point $Q(b,-\alpha t)$. The goal is then to describe the locus of points $P$, that is, to find $y$ as a function of $x$.
(a) Since car $N$ is pursuing car $M$, then at time $t$, car $N$ must be heading right at car $M$. That is, the tangent line to the curve of pursuit at $P$ must pass through the point $Q$. Using the point-slope formula, this implies:
$$y+\alpha t=\frac{dy}{dx}(x-b)$$
(1) $$\frac{dy}{dx}=\frac{y+\alpha t}{x-b}$$
(b) Since we know the speed at which car $N$ is traveling, we know the distance it travels is $\beta t$. This distance is also the length of the pursuit curve from $(0,0)$ to $(x,y)$. Using the arc length formula from calculus, we find that:
(2) $$\beta t=\int_0^x \sqrt{1+\left(y'(u) \right)^2}\,du$$
Solving for $t$ in equations (1) and (2), we find that:
(3) $$(x-b)\frac{dy}{dx}-y=\frac{\alpha}{\beta}\int_0^x \sqrt{1+\left(y'(u) \right)^2}\,du$$
(c) Differentiating both sides of (3) with respect to $x$, we find:
$$\left((x-b)\frac{d^2y}{dx^2}+\frac{dy}{dx} \right)-\frac{dy}{dx}=\frac{\alpha}{\beta}\sqrt{1+\left( \frac{dy}{dx} \right)^2}$$
Letting $$\omega=\frac{dy}{dx}$$ we obtain the first order IVP:
$$(x-b)\frac{d\omega}{dx}=\frac{\alpha}{\beta}\sqrt{1+ \omega^2}$$ where $$\omega(0)=0$$
(d) Using separation of variables, and switching the dummy variables of integration so that we may use the boundaries as the limits, we find:
$$\int_0^{\omega}\frac{1}{\sqrt{1+u^2}}\,du= \frac{\alpha}{\beta}\int_0^x\frac{1}{v-b}\,dv$$
Integrating, we find:
$$\left[\ln|u+\sqrt{1+u^2}| \right]_0^{\omega}=\frac{\alpha}{\beta}\left[\ln|v-b| \right]_0^x$$
$$\ln|\omega+\sqrt{1+\omega^2}|=\ln\left|\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}} \right|$$
Since $$-\infty\le \omega<0$$ and $$0<1-\frac{x}{b}$$ we may state:
$$\omega+\sqrt{1+\omega^2}=-\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}}$$
$$\left(\sqrt{1+\omega^2} \right)^2=\left(\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}}+\omega \right)^2$$
$$1+\omega^2=\left(1-\frac{x}{b} \right)^{\frac{2\alpha}{\beta}}+2\omega\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}}+\omega^2$$
$$2\omega\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}}=1-\left(1-\frac{x}{b} \right)^{\frac{2\alpha}{\beta}}$$
Back-substituting for $\omega$, we have the IVP:
$$\frac{dy}{dx}=\frac{1}{2}\left(\left(1-\frac{x}{b} \right)^{-\frac{\alpha}{\beta}}-\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}} \right)$$ where $$y(0)=0$$
Switching the dummy variables of integration so we can use the boundaries, we have:
$$\int_0^y\,du=\frac{1}{2}\int_0^x\left(1-\frac{v}{b} \right)^{-\frac{\alpha}{\beta}}-\left(1-\frac{v}{b} \right)^{\frac{\alpha}{\beta}}\,dv$$
$$_0^y=\frac{b}{2}\left[\frac{\left(1+\frac{v}{b} \right)^{1+\frac{\alpha}{\beta}}}{1+\frac{\alpha}{\beta}}-\frac{\left(1-\frac{v}{b} \right)^{1-\frac{\alpha}{\beta}}}{1-\frac{\alpha}{\beta}} \right]_0^x$$
After simplifying, we obtain:
$$y=\frac{b\beta}{2(\alpha^2-\beta^2)}\left((\beta-\alpha)\left(1-\frac{x}{b} \right)^{\frac{\beta+\alpha}{\beta}}-(\beta+\alpha)\left(1-\frac{x}{b} \right)^{\frac{\beta-\alpha}{\beta}}+2\alpha \right)$$
Plugging in the given data $b=36,\,\beta=2\alpha,\,x=b$ we find:
$$y=\frac{36(2\alpha)}{2(\alpha^2-(2\alpha)^2)}\left((2\alpha-\alpha)\left(1-\frac{36}{36} \right)^{\frac{2\alpha+\alpha}{2\alpha}}-(2\alpha+\alpha)\left(1-\frac{36}{36} \right)^{\frac{2\alpha-\alpha}{2\alpha}}+2\alpha \right)$$
$$y=\frac{12}{-\alpha}\left(2\alpha \right)=-24$$
Thus, the distance $d$ traveled by car $M$ is:
$$d=|y|=24$$
