2-Find The Whole Area of The Curve

 Answer:

 Here the given hyperbola is

\begin{align} \pmb{x^{2}-y^{2}=c^{2} } \end{align}

 Also given that the point \(\pmb{(a,b) }\) lies on the hyperbola \(\pmb{(1) }\). Therefore we have
\begin{align} \pmb{a^{2}-b^{2}=c^{2} } \end{align}   Now the equation of the straight line passing through from the origin to the point \(\pmb{(a,b) }\) is \begin{align} &\pmb{\frac{y-0}{x-0}=\frac{b-0}{a-0} }\nonumber\\ \implies & \pmb{\frac{y}{x}=\frac{b}{a} }\nonumber\\ \implies & \pmb{ay=bx }\nonumber\\ \implies & \pmb{x=f(y)=\frac{ay}{b} } \end{align}

  From the equation (\pmb{(1) }\), we have \begin{align} & \pmb{x^{2}-y^{2}=c^{2} }\nonumber\\ \implies & \pmb{x^{2}=c^{2}+y^{2} }\nonumber\\ \implies & \pmb{x=\pm \sqrt{c^{2}+y^{2}} }\nonumber\\ \implies & \pmb{x=g(y)= \sqrt{c^{2}+y^{2}} }\\ and~ & \pmb{x=h(y)=- \sqrt{c^{2}+y^{2}} }\\ \end{align}

  Therefore the Total Area:

\begin{align} &\pmb{ \int^{b}_{0} \left[g(y)-f(y)\right]dy }\nonumber\\ = &\pmb{ \int^{b}_{0} \left[\sqrt{c^{2}+y^{2}}-\frac{ay}{b}\right]dy }\nonumber\\ = &\pmb{\left[\frac{y\sqrt{c^{2}+y^{2}}}{2}+\frac{c^{2}}{2}\log \left(y+\sqrt{c^{2}+y^{2}}\right)-\frac{ay^{2}}{2b}\right]^{b}_{0} }\nonumber\\ = &\pmb{\frac{b\sqrt{c^{2}+b^{2}}}{2}+\frac{c^{2}}{2}\log \left(b+\sqrt{c^{2}+b^{2}}\right)-\frac{ab^{\cancel{2}}}{2\cancel{b}} -\frac{c^{2}}{2}\log c }\nonumber\\ \end{align}   From \(\pmb{(2) }\), we have \(\pmb{ a^{2}=c^{2}+b^{2} }\). Putting this value we get, \begin{align} =&\pmb{\cancel{\frac{ab}{2}}+\frac{c^{2}}{2}\log \left(b+a\right)-\cancel{\frac{ab}{2}} -\frac{c^{2}}{2}\log c }\nonumber\\ =&\pmb{\frac{c^{2}}{2}\log \left(\frac{a+b}{c}\right) }\nonumber\\ \end{align}