We can find how y and x are related by substituting t=x/(2a):
y=a(x2/(4a2)=x2/(4a), a parabola with vertex (0,0).
On the axis of symmetry (y-axis) the focus and directrix line are equidistant from the vertex. Call this distance f. All points P(x,y) on the parabola are equidistant from the focus F(0,f) and directrix line (y=-f).
PF2=(f-y)2+x2=f2-2fy+y2+x2
The y-coord of P is distance y-(-f)=y+f from the directrix line y=-f.
Since x2=4ay we can substitute:
PF=√(f2-2fy+y2+4ay).
Therefore:
√(f2-2fy+y2+4ay)=y+f,
f2-2fy+y2+4ay=y2+2fy+f2,
-2fy+4ay=2fy,
4ay=4fy⇒f=a.
The tangent at P(xP,yP) is dy/dx=xP/(2a) and at Q(xQ,yQ) dy/dx=xQ/(2a).
The tangent lines are respectively:
y-yP=(xP/(2a))(x-xP) and y-yQ=(xQ/(2a))(x-xQ).
Their intersection is given by:
(xP/(2a))(x-xP)+yP=(xQ/(2a))(x-xQ)+yQ.
(xP/(2a))(x-xP)-(xQ/(2a))(x-xQ)=yQ-yP,
x(xP-xQ)=2a(yQ-yP)+xP2-xQ2,
x=(2a(yQ-yP)+xP2-xQ2)/(xP-xQ).
yP=xP2/(4a), yQ=xQ2/(4a), 2a(yQ-yP)=½(xQ2-xP2),
x=½(xP2-xQ2)/(xP-xQ)=½(xP+xQ);
y=(xP/(2a))(x-xP)+yP,
y=(xP/(2a))(½(xP+xQ)-xP)+yP,
y=(xP/(4a))(xQ-xP)+yP=(xP/(4a))(xQ-xP)+xP2/(4a),
y=xPxQ/(4a).
This makes the intersection point X of the tangents at P and Q X(½(xP+xQ),xPxQ/(4a)).
More to follow in due course...