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January 22, 2020, 05:27:26 am

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#### spnmox

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##### parametrics question
« on: February 09, 2019, 09:41:18 pm »
0
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

#### RuiAce

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##### Re: parametrics question
« Reply #1 on: February 09, 2019, 09:58:00 pm »
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Are you sure there's no typo here? $P(2p, p^2)$ does not lie on $x^2=4ay$.

#### spnmox

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##### Re: parametrics question
« Reply #2 on: February 09, 2019, 10:02:32 pm »
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Oops, I meant x^2=4y haha.

#### spnmox

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##### Re: parametrics question
« Reply #3 on: February 09, 2019, 10:04:01 pm »
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Oh I see what happened. thanks!

#### fun_jirachi

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##### Re: parametrics question
« Reply #4 on: February 09, 2019, 11:33:14 pm »
+1
Hi!

Basically you know that that focus is (0, 1).
The perpendicular to the tangent we know has gradient -1/p (you can differentiate the curve and check.) Sub this info into the point gradient form of a line to get that x+py-p=0. This line intersects the directrix at M, so to find M, solve simultaneously ie. sub in y=-1. You get that M is the point (2p, -1). Since P is the point (2p, p^2), N is the point (2p, (p^2-1)/2). You can solve for this parametrically to get that the locus of N is the parabola x^2=8(y+1/2) which is just a parabola with vertex (0, -1/2), focal length 2.

Hope this helps
« Last Edit: February 09, 2019, 11:35:56 pm by fun_jirachi »
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