x=4ay2,
k=xy′=2ay,
x−ati2=2a2ati(y−2ati),
{x−at12=t1(y−2at1),x−at22=t2(y−2at2);
{x0=at1t2,y0=a(t1+t2); – coordinates of point P.
Line PM (parallel to the x-axis) has an equation y=a(t1+t2).
Midpoint of AB has coordinates (2a(t12+t22), a(t1+t2)).
AB equation: 2x−(t1+t2)y+2at1t2=0,
with a PM equation we'll have
{x0=2a(t12+t22),y0=a(t1+t2); – coordinates of point M, and they are equal to the coordinates of point P.
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