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parametrics help please! (1 Viewer)

sean182

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i need help please anyone!

P1, P2 are points on parabola x^2 = 4ay with parameters p and 1/p. if the tangents at P1 and P2 intersect at R, prove the locus of are is the line y=a.
 

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Put in parametric form, P1(2ap,ap2), P2(2a/p,a/p2). Since the tangent to the parabola at a point (2at,at2) has gradient t, the parameter at the point, and use point-gradient form, the equations of the tangents at P1 and P2 are y=px-ap2 and y=x/p-a/p2. Solve simultaneously, one obtains R(a(p+1/p),a). Therefore, regardless of the value of p, the y-coordinate of R is always a. Thus, the locus of R is the line y=a.
 

GaDaMIt

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sean182 said:
i need help please anyone!

P1, P2 are points on parabola x^2 = 4ay with parameters p and 1/p. if the tangents at P1 and P2 intersect at R, prove the locus of are is the line y=a.
you mean R?
 

falcon07

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The two points are:

P1 (2ap, ap^2)
P2 (2a/p, a/p^2)

At p1, x = 2ap, y = ap^2

dy/dp = 2ap
dx/dp = 2a
dy/dx = dy/dp . dp/dx = 2ap x 1/2a = p (gradient of tangent at P1)
Similarly, gradient of tangent at P2 = 1/p

Point gradient formula:
Tangent at P1:
y - ap^2 = xp - 2ap^2
y = xp - ap^2 (1)

Tangent at P2:
y - a/p^2 = x/p - 2a/p^2
y = x/p - a/p^2 (2)

Intersection of the 2 lines
xp - ap^2 = x/p - a/p^2
xp^3 + xp = ap^4 - a (multiply by p^2)
xp(p^2 + 1) = a(p^4 - 1) = a(p^2 + 1)(p^2 - 1)
x = a/p . (p^2 + 1)

Sub back into (1)
y = ap/p (p^2 + 1) - ap^2
y = ap^2 + a - ap^2
Hence y = a

R ( a/p . (p^2 + 1) , a)
Therefore the locus of point R is y = a, as a is a constant.
 

sean182

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whao this place is great, thanks alot. i was seriously working on that question for an hour.
 
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pLuvia

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Post some questions then people can assist you with your problems
 

Riviet

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テリー said:
ok ><
im having a great tragedy with Parametrics T.T
can some1 give me some tips on it
i fully understnd the theories but i just cant solve the problems..
advice please
thx in advance
Often it helps if you draw a diagram, labelling as much as you can with whatever info you know. After doing that, work out what you need to find and work away. Hope that helps.
 

テリー

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the thing is every single question i do i seem to get stuck, i hav no idea how to find the things the question ask me to find.... ><
for example (extract from fitzpatrick green book 24c)

Q15. Find the co-ordinates of three points on the parabola x^2=4y such that the normals through these points pass through the point (-12,15)

etc i cant do a SINGLE question... i mean in what way do i need to think of the question to realize that i hav to do this and that to get the answer etc

and Riviet i do draw diagrams, put in important properties of the question, jst doesnt help my brain cells
 

Raginsheep

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For Q15, you should thing about what exactally the question is asking you. If it helps, underline the key words, information and write down everything that question asks you to find. In this case, its about normals that pass through a specific point.

Realising that the question involves normals, you should get the idea that tangents would also probably be invloved. So differentiate x2 = 4y.
d/dx x2/4 = x/2.
This means that the gradient of the tangent is x/2 which means the graident of the normal is -2/x.

Sub this into the linear form of a line to get y-y0 = -2/x (x- x0). Simplified, this will give 2x + xy -xy0 = 2x0. Sub in (-12, 15) to get 13x - xy = 24 which you feel might be a bit tricky to solve with the 'xy' term. However, since this is parametrics which means that we can turn a problem of two variables into a problem of one, we sub in (2p, p2) to get p3 - 13p + 12 = 0 which you can solve.

Since its a quartic, there should be three roots, and given the question, three distince roots or values for p. This since you have the three values of p, just sub them into the general parametric definition of a parabola (2p, p2) to get your three points.

Hope that makes sense....
 

Riviet

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Just a minor correction - it's a cubic equation, not a quartic, that gives the three roots.
 

cocopuk

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hi guys, i'm only just having major problems with parametrics and locus also an exam is coming up pretty damn soon (monday) and i cant even do the first question of the parametric & locus exercise. Here's the question hope you guys can help =D and thanks in advance for your time and effort XD ehehe

from a point P on the prabola x^2=2y, a tangent is drawn. From the focus S, a perpendicular is drawn to meet the tangent at R.
(a) find the equation of SR
(b) find the locus of R.

i understand the question, its mainly the setting/working out wich confuses (especially when i get the wrong answer)
 

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