At a point p on the parabola x^2=4ay a normal pk is drawn. from vertex o, a perpendicular om is drawn to meet the normal at m. show that the equation of the locus of m as p varies on the parabola is x^4-2ax^2y+x^2y^2-ay^3=0

Its too long to write here, so I will just state what I did. I let P=(2ap,ap^2) and Q=(2aq,aq^2) But x-coordinates of P and Q differ by (2a) So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2) So Q=(2a(p-1), aq^2) which means, 2aq = 2a(p-1) therefore, q=p-1 then I subbed that value of q in aq^2 so Q=(2a(p-1), a(p-1)^2) and P=(2ap,ap^2) Using these two values, I found the midpoint which was: M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 ) then x = a(2p-1) rearranging to make p the subject p= (x+a)/2a

That’s what i got when i looked up what the solutions were