Given curve is y2=4(x−2) and line y=2x−8
Finding the point of intersection of y=2x−4 and y2=4x.
⇒x=2y+8=4y2+8
⇒2y+16=y2+8
⇒y2−2y−8=0
⇒(y−4)(y+2)=0
⇒y=4,−2
⇒x=6,3

Area of the bounded region is given by, A=∫−24(2y+8−(4y2+2))dy
⇒A=[4y2+2y−12y3]−24
⇒A=4+8−316−1+4−32
⇒A=9sq.units.