SOLUTION: A point P(2p,p^2) lie on x^2=4ay. The perpendicular, from the focus S of the parabola, on to the tangent cuts the directrix at M. If N is the midpoint of the interval PM, find the

Question 1134394: A point P(2p,p^2) lie on x^2=4ay. The perpendicular, from the focus S of the parabola, on to the tangent cuts the directrix at M. If N is the midpoint of the interval PM, find the equation of the locus of N.
I keep getting long equations that I don't know how to simplify. Right now I have this: 2y+4ay+2a^y=x^2-a-2a^2-a^3
Answer by t0hierry(194) (Show Source): You can put this solution on YOUR website!
I don't know how to solve your problem. All I can do is write down for you the equation of a parabola.
There are two kinds, vertical and horizontal. IF the parabola is vertical, along y, then it is of the form {{y = Ax^2 + B}}.
You derive this by recalling that the distance to the focus is equal to the distance to the directrix.
{{(x - a)^2 + (y-b)^2 = (y-k)^2}}
Here the focus is a point of coordinates (a,b). And the directrix has equation y = k.
I think the equation you're looking for is similar to this one, which is why I go through its steps:
{{y^2 - 2ky + k^2 = y^2 -2by + b^2 + (x-a)^2}}
y^2 cancels out, as expected, and we end up with
{{y(2b - 2k) = b^2 - k^2 + (x-a)^2}} or
{{y = (x-a)^2/[2(b-k)] + (b+k)/2}}
The other equation for the parabola, has a physical explanation. It's the trajectory of a point mass, dropped from a plane, or shot through a canon. There is no acceleration in the horizontal direction, so v = constant.
And thus x = vt.
At the same time, there is acceleration along the y direction, that of gravity. And naturally you have v = gt and {{y = 1/2 g t^2 =1/2 g(x/v)^2 = 1/2 g/v^2 x^2}}
I was hoping someone else solved this problem for you but, if they did, I missed it. Anyway, I hope this helps.
RELATED QUESTIONS
Does (1,11) lie on the parabola defined by the focus (0,4) and the directrix y =... (answered by greenestamps)

A parabola has a directrix of y=-3 and a focus of (0,5). Show that the point (12,10) must (answered by josgarithmetic)

P (2p,p^2) is a variable point on the parabola x^2 = 4y. N is the foot of the... (answered by KMST)

Verify that the point A = (8, 25/3) lies on the parabola whose focus is (0, 6) and whose... (answered by greenestamps)

Write an equation that says that P = (x, y) is on the parabola whose focus is (2, 1) and (answered by josgarithmetic)

Let λ be the line y = 1 and F be the point (−1,2). Verify that the point (2,6) is... (answered by greenestamps)

1. Use the focus-directrix definition of a parabola to answer the following questions. (answered by KMST)

Draw the parabola. Identify the focus and directrix and show them on the graph.... (answered by lwsshak3)

Hello tutor, This one question on my online homework assignment has been puzzling me for (answered by lwsshak3)