derivatives
The line \begin{equation*}y=m(x-m)\end{equation*} is tangent to the curve \begin{equation*}
(1-2x)y=1
\end{equation*}
How do you go about finding the coordinates of the point where they intersect? I differentiated the curve and got \begin{equation*}\frac{dy}{dx}=\frac{2}{(1-2x)^2}\end{equation*}
How do I proceed on from this? Do I just equate the derivative to m and plot a graph using Wolfram?
If you want a horizontal tangent, then you need to look for where ${dy\over dx}=0$.
Edit: As indicated, solve ${dy\over dx}=0\implies y(2.5-1/x)=0\implies y=0\text{ or }x={2\over 5}$. But $y=0$ is not in the domain of the original relation, $\ln(xy)=2.5x$, so the horizontal tangent occurs only when $x=2/5\implies y=5e/2$.
This is depicted graphically below: the dashed line is the horizontal tangent passing thru $(2/5,5e/2)$ on the graph of $\ln(xy)=2.5x$.
Equation of tangent at point $(x_0, f(x_0))$ is $$y = f(x_0) + f'(x_0) \cdot (x-x_0)$$
Best Answer
Suppose the line $y=m(x-m)$ touches the curve at the point $(a,b)$. Then the slope of the tangent at $(a,b)$ to the curve $(1-2x)y=1$ must be $m$ as the slope of the line $y=m(x-m)$ is $m$. So by your calculation $$\frac{2}{(1-2a)^2}=m.$$ But $(a,b)$ is a point on the line $y=m(x-m)$, whence $b=m(a-m)$. Now you have two unknowns and two equations. So find $(a,b)$.