A point $P (\frac{e^t+e^{-t}}2,\frac{e^t-e^{-t}}2)$ traces a locus $S=0$ in $XY$, a fixed point $P'$ having parameter $t'$ lies on $S=0$. Area bound by the curve, the line $OP'$ ($O$ being origin) and $x$-axis is $240$ sq units. Find value of $t'$. " t is any non negative parameter
Now here on finding the locus traced of S=0, It's an hyperbola $x^2-y^2=1$.
$$x=(e^{t}+e^{-t})/2, y=(e^{t}-e^{-t})/2$$ Squaring and subtracting, we get $x^2-y^2=1$, which represents a hyperbola.
First of all $\cosh{t}\geq 1$ so $(x,y)=(\cosh{t},\sinh{t})$ is the right branch of the hyperbola $x^2-y^2=1.$ Call the $x$-coordinate of $P'$ $Q'.$ The area under the hyperbola above the $x$-axis is $A_1=\int_0^{t'} y(t) x'(t) dt=\int_0^{t'}\sinh^2{t} \, dt.$ So $\operatorname{area}(\triangle OP'Q')-A_1=240$ or $\frac12 \cosh{t'}\sinh{t'}-(\frac14 (\sinh{2 t'} - 2 t'))=240,$ this has solution $t'=480$ since $\sinh{2 t'}=2\sinh{t'}\cosh{t'}.$
