[HELP] parametric equation Q:

[ryan882]

A Parabola is given parametrically by x=at^2, y=2at. If P is any point on the parabola, let F be the foot of the perpendicular from P onto the x axis of symmetry. Let G be the point where the normal from P crosses the axis of symmetry.
Prove that FG= 2a
??

 

[HallsofIvy]

A Parabola is given parametrically by x=at^2, y=2at. If P is any point on the parabola, let F be the foot of the perpendicular from P onto the x axis of symmetry.

So F is (at^2, 0).

Let G be the point where the normal from P crosses the axis of symmetry.

dx/dt= 2at and dy/dt= 2a so dy/dx= 1/t. The normal to the parabola at P= (at^2, 2at) has slope -t and so y= (-t)(x- at^2)+ (2at) where "t" is the specific value at P. That line crosses the x- axis when y= (-t)(x- at^2)+ (2at)= 0 or x= at^2+ 2a. G is (at^2+ 2a, 0).

Prove that FG= 2a
??