What is the maximal area between hyperbola and chord $AB$? 
Suppose $A$ and $B$ are variable points on upper branch of hyperbola $\mathcal{H}:\;x^2-y^2=-4$ such that $AB = 1$. What is the maximal area between $AB$ and $\mathcal{H}$?


Clearly $y= \sqrt{x^2+4}$. If we set $A\big(t,\sqrt{t^2+4}\big)$ then $B\big(t+h, \sqrt{(t+h)^2+4} \big)$ for some positive $h$, such that $$h^2 +\Big(\sqrt{(t+h)^2+4} -\sqrt{t^2+4} \Big)^2=1$$
and we have to find a minimum value of
$$f(t) = \int_{t}^{t+h}\sqrt{x^2+4}\;dx$$
Notice that if $$F(x):= \int\sqrt{x^2+4}\;dx$$   then $$ F(x)= 2\ln\left(\left|\sqrt{x^2+4}+x\right|\right)+x\sqrt{\dfrac{x^2}{4}+1}$$
and things get complicated very quickly.

Disclaimer: I am not sure since I am still new to calculus.


Find the minimum value of $f(t) = \displaystyle\int_{t}^{t + h}\sqrt{x^{2} + 4}\,dx$.

The fundamental theorem of calculus can be applied here:
\begin{align*}\dfrac{d}{dt}f(t) &= \dfrac{d}{dt}\int_{t}^{t + h}\sqrt{x^{2} + 4}\,dx \\ f'(t) &=\left(\sqrt{(t + h)^{2} + 4}\right)\left(\dfrac{d}{dt}(t + h)\right) - \left(\sqrt{t^{2} + 4}\right)\left(\dfrac{d}{dt}(t)\right) \\f'(t) &= \left(\sqrt{(t + h)^{2} + 4}\right) - \left(\sqrt{t^{2} + 4}\right)\end{align*}
Then, set $f'(t) = 0$,
\begin{align*}f'(t) &= 0 \\ \sqrt{(t + h)^{2} + 4} - \sqrt{t^{2} + 4} &= 0 \\ \sqrt{(t + h)^{2} + 4} &= \sqrt{t^{2} + 4} \\ (t + h)^{2} + 4 &=t^{2} + 4 \\ t^{2} + 2th + h^{2} + 4 &= t^{2} + 4 \\ 2th + h^{2} &= 0 \\ 2t &= -h \\ t&= -\dfrac{h}{2}\end{align*}
I don't know how to proceed now, but I hope that this can help.

I used the variant of the FTOC which is: $$\dfrac{d}{dx}\int_{u(x)}^{v(x)}f(t)\,dt = f(v(x))v'(x) - f(u(x))u'(x)$$
