ok so choose a point on the parabola x^2=4ay: P(2ap,ap^2).
Now let's find the normal at that point.
y=x^2/4a
dy/dx=2x/4a
x=2ap, dy/dx=p so gradient of normal is -1/p
y-ap^2=-1/p (x-2ap)
Sub (0, Ka) into the equation[since you are given that the normals pass through (0,Ka)]
Ka-ap^2= (-1/p)(-2ap)
= 2a
Ka = 2a + ap^2
K = 2 + p^2
p^2= k-2
p = -√(k-2) or √(k-2) where k>2
From this, we get two solutions for the normal but this does not account for the fact that there is another normal at the origin (where p=0) which is the axis of the parabola and also passes through (0, Ka) so there is actually 3 normals which pass (0, Ka).
For the second part, just sub k=3 into p = -√(k-2) or √(k-2) so you get: p= -1 or 1. Also remember that p=0 is another solution since there is a normal which passes through the origin and (0,3a). Sub those 3 values into P(2ap,ap^2) and you will get your 3 points
I hope this makes sense
