Finding the distance between two parallel tangents of a rational function A Curve C has equation $y=\frac{2x-5}{x-1}$
a) A line y = mx+c is tangent to the curve. Find a condition for c in terms of m.
This can be solved relatively easily
$mx + c = \frac{2x-5}{x-1}$
$mx^2 + (c-m-2)x -c+5=0$
take discriminant as 0, as there is only one point at which the tangent intersects the curve
$c^2 +(2m-4)c +m^2 -16m+4=0$
By completing the square for c and m separately
$(c+m-2)^2-(m-2)^2+(m-8)^2-64+4=0$
$(c+m-2)^2 = 12m$
$c = 2-m\pm\sqrt{12m}$
b) Two parallel lines are each tangents to the curve, touching at points P(p,q) and R(r,s) with p<1

I have plotted this graph using desmos and attached an image of it
P can be taken to be a point in the top-left part of the graph and R a point in the bottom-right section of the graph.
If the distance $PR$ is $d$, find $d^2$ in terms of $m$, the gradient of each of the lines.
No calculus is required for this part.
$d^2=(p-r)^2 +(q-s)^2$ 
the tangents intersecting P and Q are given by:
$y = mx - mp + q$
$y = mx - mr +s$
$PR$ is perpendicular to the tangents
Therefore $grad_{PR} = -\frac{1}{m}$
Using the original equation:
$q = \frac{2p-5}{p-1}$
and $s = \frac{2r-5}{r-1}$
$q-s = \frac{3(p-r)}{(p-1)(r-1)}$
also take the leftmost tangent as $y_1$ and the other as $y_2$
$y_1 = mx - mp + \frac{2p-5}{p-1}$
$y_2 = mx - mr + \frac{2r-5}{r-1}$
$grad_{PR}=-\frac{1}{m} =\frac{p-r}{q-s} = \frac{(p-1)(r-1)}{3}$
$m = -\frac{3}{(p-1)(r-1)}$
$(q-s)^2 = \frac{9(p-r)^2}{(p-1)^2(r-1)^2}$
$(p-r)^2=\frac{(p-r)^2(p-1)^2(r-1)^2}{(p-1)^2(r-1)^2}$
$d^2 = \frac{(p-r)^2[9+(p-1)^2(r-1)^2]}{(p-1)^2(r-1)^2}$
$(p-1)^2(r-1)^2 = \frac{9}{m^2}$
$d^2 = (p-r)^2(m^2+1)$
using the conclusion from part a
$y_1  = mx + 2 - m - \sqrt{12m}$ 
$y_2 = mx+ 2- m + \sqrt{12m}$
Hence $ 2 - m - \sqrt{12m} = - mp + \frac{2p-5}{p-1}$
and $ 2- m + \sqrt{12m} =- mr + \frac{2r-5}{r-1} $
I still need to find $p-r$ in order to express $d^2$ in terms of $m$
I've been working on this problem for ages so any help would be much appreciated
The answer is 
$d^2 = 4(\frac{3}{m}+3m)$
however, this is not particularly useful without a solution to it
c) By differentiating $d^2$ with respect to $m$, or otherwise, find the shortest distance between the two branches of the curve.
This part evidently does require calculus.
Any suggestions as to how I should answer this question would also be helpful.

 A: Given $y=\frac{2x-5}{x-1}$, the two tangents must have same slope:
$$m=y'(p)=\frac{3}{(p-1)^2}=y'(r)=\frac{3}{(r-1)^2} \Rightarrow p=-\sqrt{\frac{3}{m}}+1; r=\sqrt{\frac{3}{m}}+1 \ \ \ (*)$$
The squared distance between the tangents is
$$d^2=(p-r)^2+(q-s)^2=(p-r)^2+\left(\frac{2p-5}{p-1}-\frac{2r-5}{r-1}\right)^2.$$
Now, it remains to plug (*) in here.
Note: $p<1$.
A: Here’s a somewhat different approach using homogeneous coordinates.  
The graph of this rational function is a hyperbola, so rewrite the equation as $$xy-2x-y+5 = 0$$ and then put it in the matrix form $\mathbf x^TC\mathbf x=0$, with $$C=\begin{bmatrix}0&\frac12&-1\\\frac12&0&-\frac12\\-1&-\frac12&5\end{bmatrix}.$$ Tangent lines $\mathbf l$ to this conic satisfy the dual conic equation $\mathbf l^TC^{-1}\mathbf l=0$. Setting $\mathbf l=(m,-1,c)^T$, corresponding to the generic equation $mx-y+c=0$ of a line, gives the equation $$\begin{bmatrix}m&-1&c\end{bmatrix} \begin{bmatrix}\frac13&\frac83&\frac13\\\frac83&\frac43&\frac23\\\frac13&\frac23&\frac13\end{bmatrix} \begin{bmatrix}m\\-1\\c\end{bmatrix} = \frac13\left(c^2+(2m-4)c+(m^2-16m-4)\right) = 0$$ with solutions $$c = 2-m\pm2\sqrt{3m}.$$ The point of tangency of a tangent line $\mathbf l$ is its pole $C^{-1}\mathbf l$: $$\begin{bmatrix}\frac13&\frac83&\frac13\\\frac83&\frac43&\frac23\\\frac13&\frac23&\frac13\end{bmatrix} \begin{bmatrix}m \\ -1 \\ 2-m\pm2\sqrt{3m} \end{bmatrix} = \begin{bmatrix} -2\pm2\sqrt{\frac m3} \\ 2m\pm4\sqrt{\frac m3} \\ \pm2\sqrt{\frac m3}\end{bmatrix}$$ which when converted back into inhomogeneous Cartesian coordinates is $$\left( 1\mp\sqrt{\frac3m}, 2\pm\sqrt{3m} \right).$$ The first choice of signs gives the point $P$, the other $R$.  
Since the problem you’re working on says, “or otherwise,” here’s an otherwise. The curve is a hyperbola, so the minimum distance between its branches occurs at the vertices, and this distance is just the hyperbola’s transverse axis length. This can be computed directly from the matrix $C$: it is equal to $$2\sqrt{{-\det C \over \lambda_+^2\lambda_-}},$$ where $\lambda_+$ and $\lambda_-$ are the positive and negative eigenvalues of the upper-left $2\times2$ submatrix of $C$. This mysterious formula is the end-result of converting the general conic equation into standard form. The eigenvalues are easily seen to be $\pm\frac12$, and a simple computation then gives a minimal distance of $2\sqrt6$.  
You could also find this distance by finding the hyperbola’s vertices, which are the intersections of the hyperbola with its transverse axis. The hyperbola’s center lies at the intersection of its asymptotes, here $(1,2)$, and its transverse axis is a bisector of the asymptotes, so the equation of its transverse axis is $(x-1)+(y-2)=0$, or $x+y=3$. Solving the resulting system of equations produces the points $(1\pm\sqrt3,2\mp\sqrt3)$ and the distance between them is again $2\sqrt6$.
