AP EAPCET 23-Aug-2021 Shift 1 Paper

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