Given parabola, we have x2=4ay and y2=4ax Intersection point of these two parabolas are (0,0) and (4a,4a). Given that, 2bx+3cy+4d=0 passes through point of intersection. Case I If it passes through (0,0), we obtain 2b(0)+3c(0)+4d=0 ⇒d=0 ...(i) Case II If it passes through (4a,4a), we obtain 2b(4a)+3c(4a)+4(0)=0⇒(2b+3c)(4a)=0 ⇒2b+3c=0 ...(ii) From Eqs. (i) and (ii), we get d2+(2b+3c)2=0