Area of the triangle formed by the line $$x + y= 3$$   and angle bisectors of the pair of straight lines $${x^2} - {y^2} + 2y = 1$$    is-

Solution :
$$\eqalign{ & {x^2} - {y^2} + 2y = 1 \cr & \Rightarrow x = \pm \left( {y - 1} \right) \cr} $$

Bisectors of above lines are $$x = 0$$  and $$y= 1$$

So area between $$x = 0, \,y= 1$$    and $$x + y =3$$   is shaded region shown in figure.
Area $$ = \frac{1}{2} \times 2 \times 2 = 2\,{\text{sq}}{\text{.}}\,{\text{units}}$$