If a triangle is formed by any three tangents of a parabola Y2 = 4ax, two of whose vertices lie on the parabola x2 = 4by, then find the locus of the third vertex.

Solution please?

let y=mx+c  be a tangent. y^2 =4ax. solving the previous two equations simultaneously, a=mc. thus, the three tangents are

y=ax/c1 +c1, y= ax/c2 +c2, y=ax/c3+c3.   let the three vertices be (x12,y12), (x13,y13) and (x23,y23).xij = cicj/a and yij=ci+cj,  (x12)^2 = 4by12, and

(x13)^2 =4by13. thus, (c2/c3)^2 = (c1 +c2)/(c1+c3), which implies that c2c3=0. thus x23=c2c3/a=0 is the locus, which is nothing but the y-axis