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I have two equations: $$f(x) = x^2/k^2$$ and $$ g(x) = a \cdot sinh(bx)$$ The derivatives are therefore $$ f'(x) = 2x/k^2$$ and $$g'(x) = ab \cdot cosh(bx)$$

Now for given values of $k$ and $x$ (for example: $k=0.42$ and $x=7$) I want to find values of $a$ and $b$ such that the values and also the slopes of $f$ and $g$ are identical, which should be doable by solving the set of two equations $f=g$ and $f'=g'$ simultaneously.

From equation $f$ I can get $$ a= \frac{x^2}{k^2 \cdot sinh(bx)}$$ and insert that in $g$, or from $g$ I can get $$ a= \frac{2x}{bk^2 \cdot cosh(bx)}$$ and insert that in equation $f$, but then I am still not able to solve the resulting single equation.

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    i think it can only be done by an numerical way – 2017-01-16

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Since you have $a$ in both equation, you can eliminate it and the equation becomes $$ \frac{x}{\sinh(bx)}=\frac{2}{b\cosh(bx)} $$ that is, $$ bx=2\tanh(bx) $$

Consider the function $$ f(t)=2\tanh t-t $$ that we can study for $t\ge0$. The derivative is $$ f'(t)=\frac{2}{\cosh^2t}-1=\frac{2-\cosh^2t}{\cosh^2t} $$ which vanishes at $\cosh t=\sqrt{2}$ and is positive in the interval $[0,\operatorname{arcosh}\sqrt{2}]$. Moreover, $\lim_{t\to\infty}f(t)=-\infty$. Thus the equation $f(t)=0$ has a single positive solution $t_0$, that can be determined numerically; an approximate value is $1.9189$.

Once you have found it, you get $$ b=\frac{t_0}{x} $$ and you can compute the value of $a$.

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    Very helpful, thank you very much. Using Wolfram Alpha, I get a value $t_0 \approx 1.9150080481545$. – 2017-01-17