<s>[INST] 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.
 [/INST] Parametrize the hyperbola as $x=2\sinh t $, $y= 2\cosh t$  and let the endpoints of the chord be $A(2\sinh a, 2\cosh a)$, $B(2\sinh b, 2\cosh b)$. Then, the equation of chord $AB$ is
$$y= x\tanh\frac{a+b}2 +2\frac{\cosh \frac{a-b}2}{\cosh\frac{a+b}2 }$$
and the area between the chord and hyperbola is
\begin{align}
K =&\int_{x_a}^{x_b}( y - \sqrt{x^2+4})dx\\
=&\int_a^b \left(2\tanh\frac{a+b}2 \sinh t +  
 2\frac{\cosh \frac{a-b}2}{\cosh\frac{a+b}2 }
 - 2 \cosh t  \right)2\cosh t\>dt\\
=& 2\tanh\frac{a+b}2 (\sinh^2 b -\sinh^2a)
+2\sinh(b-a)(2-\cosh (a+b) )-2(b-a)
\end{align}
Moreover, $1=AB^2=(x_b-x_a)^2 + (y_b-y_a)^2 $ leads to
$$\cosh2b +\cosh 2a -2 \cos(a+b)=\frac14$$
$$\frac{d b}{da}
=\frac{\sinh2a}{\sinh(a+b)-\sinh2b}$$
Then, it is straightforward to verify that $\frac{d b}{da} = 1 $, $\frac{d K}{da} = 0 $ if $b=-a$, which leads to $\sinh b =-\sinh a=  \frac14$ and the maximum area
$$K_{max}= \frac{\sqrt{17}}4-4\sinh^{-1}\frac14$$
</s>