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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.
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