Touch 'n Go

We can simplify the equation of the curve to one that we can deal with much easier.

$\begin{aligned} \frac{x-y+7}{7-y}+\frac{5x+7}{5x+8}&=2\\ \frac{x}{7-y}+1+\frac{5x+7}{5x+8}&=2\\ \frac{x}{7-y}=1-\frac{5x+7}{5x+8}&\\ \frac{x}{7-y}=\frac{1}{5x+8}&\\ 7-y=5x^{2}+8x&\\ y=-5x^{2}-8x+7&\\ \end{aligned}$

The slope of the tangent line to this parabola is given by its derivative, $-10x-8$ .

There must be two points on the parabola whose tangent lines have a $y$ -intercept of $10$ .

Let's say that the $x$ -coordinate of one of these two points is $a$ .

$\begin{aligned} mx+10&=(-10a-8)x+10\\ &=-10ax-8x+10\\ \end{aligned}$

$\begin{aligned} -10ax-8x+10&=(-10a-8)(x-a)+(-5a^{2}-8a+7)\\ -10ax-8x+10&=-10ax+10a^{2}-8x+8a-5a^{2}-8a+7\\ 5a^{2}&=3\\ a&=\sqrt{\frac{3}{5}}\\ a&=-\sqrt{\frac{3}{5}}\\ \end{aligned}$

At these two values of $a$ , the corresponding $y$ -values on the parabola are $-8\sqrt{\frac{3}{5}}+4$ and $8\sqrt{\frac{3}{5}}+4$ respectively.

This makes $x_{1}\times y_{2}+x_{2}\times y_{1}=\frac{m}{n}=\frac{48}{5}$ , so $m+n=53$ .