Find a and b such that the line $10ex + 10y = 0$ will be a tangent of the curve $ y = ae^{1/(x-1)} - b$ at $x = 2$?

Question

How do I solve this, it seems very simple, yet when I try it turns impossible because I have no idea how to continue: $$ 10ex+10y=0 \\ ex+y=0\\ y=-ex\\ f(x) = -ex\\ f(2)=-2e $$ And now? I am not sure if this is the right path to solving this so far, and as for the next steps I think I need to find the derivative of the 2nd line. I end up with $$y =- ae^{\frac{1}{x-1}}\frac{1}{(x-1)^2} $$ and I have no idea how to continue from here, I think I am supposed to use $-2e$ somewhere, but I don't understand how that will help me find $a$ or $b$, considering $b$ is gone now?

Answer 1

Let $$f(x)=ae^{\frac{1}{x-1}}-b$$

then for $x>1$ $$f'(x)=-\frac{1}{(x-1)^2}(f(x)+b)$$

the equation of the tangent at $x=2$ is

$$y=f'(2)(x-2)+f(2)$$ $$=-(f(2)+b)(x-2)+f(2)$$ $$=-(f(2)+b)x+3(b+f(2))-b$$ $$=-ex$$ thus

$$f(2)+b=ae=e$$ and $$3ae-b=0$$ therefore $$a=1\;;\;b=3e$$