The area $A$ of the shaded region in the figure is $A=\frac{t}{2} .$ Prove this as follows:
(a) Let $A^{*}$ be twice the area outlined in the figure, and $(x, y)$ be the coordinates of $P$. Explain why
$$
A^{*}=x \sqrt{x^{2}-1}-2 \int_{1}^{x} \sqrt{u^{2}-1} d u
$$
(b) Differentiate $A^{*}$ to show that $\frac{d A^{*}}{d x}=\frac{1}{\sqrt{x^{2}-1}}$
(c) Show that $A^{*}=\cosh ^{-1} x+C,$ and explain why $C=0$.
(d) Why does it follow that $A^{*}=t ?$