Given equation of lines are (a3+3)x+ay+a−3=0 and (a5+2)x+(a+2)y+2a+3=0( a real ) Since point of intersection of lines lies on y-axis. ∴ Put x=0 in each equation, we get ay+a−3=0 and (a+2)y+2a+3=0 On solving these we get (a+2)(a−3)−a(2a+3)=0 ⇒a2−a−6−2a2−3a=0 ⇒−a2−4a−6=0⇒a2+4a+6=0 ⇒a=
−4±√16−24
2
=
−4±√−8
2
(not real) This shows that the point of intersection of the lines lies on the y -axis for no value of ' a '.