<s>[INST] 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.

 [/INST] 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: $p<1$.
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