I am trying to maximize the size of a trapezoid inscribed in an hyperbola as a function of parameter $t$. For simplicity, assume hyperbola parameters are > 0 and I'm only interested in the y$\geq$0 sheet. The upper line of the trapezoid is fixed - defined by the point $(x_{1}, y_{1})$.
For a parameterized hyperbola x(t) = $b * sinh(t)$, y(t) = $a * cosh(t)$.
The width of the upper line, $w_{1}$, is a constant: $2 * x_{1}$.
The width of the lower line, $w(t) = 2 * x(t)$ or $2 * b * sinh(t)$.
The height, $h(t) = y_{1} - y(t)$ or $y_{1} - (a*cosh(t))$.
Trapezoid area $A$, is $1/2 * (w_{1} + w(t)) * h(t)$. So it should be 'simple' to substitute in identities, take the derivative w.r.t. $t$, set to $0$, and solve for $t$. But no matter what I do, the derivative I derive is unsolvable or evaluates only to $t=0$. E.g. if I break the area into $1/2 * (w_{1} * h(t) + 1/2 * (w(t) * h(t)$, I end up with a sum including a $sinh(t)$ term, a $cosh(t)$ term, and $sinh(t)*cosh(t)$ term and the derivative of that last term is what seems to mess things up.
For comparison, I easily solved a similar problem for a parabola with standard differential calculus - I expressed the area of the trapezoid as a function of a parameter, took the derivative and set it to 0 and solved for the parameter.
We need to maximize $$\mathcal A=\frac{w_1+w(t)}{2}h(t)=(x+x_1)(y_1-y)=ab(\sinh(t)+\sinh(t_1))(\cosh(t_1)-\cosh(t)),$$ where in the last step we used the fact that since $x_1$ and $y_1$ lie on the hyperbola, then $(x_1,y_1)=(a\sinh(t_1),b\cosh(t_1)))$.
We can now differentiate wrt to $t$ and impose it to be $0$: $$ab[\cosh(t)(\cosh(t_1)-\cosh(t))-\sinh(t)(\sinh(t)+\sinh(t_1))]\overset{!}{=}0$$ $$\sinh^2(t)+\cosh^2(t)-[\cosh(t)\cosh(t_1)-\sinh(t)\sinh(t_1)]=0$$ $$\cosh(2t)=\cosh(t-t_1)$$ Therefore we have two solutions: $$\left\{ \begin{aligned} &2t=t-t_1\implies t=-t_1\\ &2t=t_1-t\implies t=\frac{t_1}{3} \end{aligned} \right.$$ It can then be checked using $2^{\text{nd}}$ derivative test or first derivative limit from the left and right of a critical point, that we can distinguish the solution that yields maximum area: $$\mathcal A'(-t_1^-)<0\ \text{and }\ \mathcal A'(-t_1^+)>0\implies -t_1\text{ is a minimum}$$ $$\mathcal A'((t_1/3)^-)<0\ \text{and }\ \mathcal A'((t_1/3)^+)<0\implies \frac{t_1}{3}\text{ is a maximum}$$
Thus, the maximum area of a trapezoid bounded by a hyperbola and $x_1$ and $y_1$ is obtained when its small base has coordinates $(x_{\text{max}},y_{\text{max}})=(a\sinh(t_1/3),b\cosh(t_1/3))$. Maybe you could express the solution in terms of $x_1$ and $y_1$, but I'll leave it as an exercise to the reader as you have to use third angle formula for hyperbolic functions.