ATAR Notes: Forum
HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: spnmox on February 09, 2019, 09:41:18 pm
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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
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Are you sure there's no typo here? \(P(2p, p^2)\) does not lie on \(x^2=4ay\).
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Oops, I meant x^2=4y haha.
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Oh I see what happened. thanks!
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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 :)