P(2ap, ap²) is a point on the parabola 4ay=x². The normal at P to the parab meets the y-axis at R. Q is the...

foot of the perpendicular from P onto the y-axis. Show that QR is a constant.

My understanding is that I have to find the points Q and R by finding the y-int. of normal at P and the y-int of the line PQ, and then use the distance formula to find PQ and hence prove that it's a constant.

I figured the points to be R(0, 2a+ap²) and Q(0, ap²). I used the distant formula, but it's not getting me anywhere so I'm assuming I'm approaching the question wrongly. Can anyone please help?

I wonder if you know how really close you came. Everything you did is correct. You don't need the distance formula at all. QR is just 2a + ap^2 - ap^2 which equals 2a. In other words, 2a is the distance between the two y values.

a is a constant for every point. So is 2a. The question is done. Way to go. This is was not an easy question.