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Locus parametrics question (1 Viewer)

sadpwner

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How do you solve part ii of this question?

Q(2aq,aq^2) is a fixed point on the parabola x^2=4ay where a>0. P(2ap,ap^2) and R(2ar,ar^2) are variable points which move on the parabola such that the chord PR is parallel to the tangent to the parabola at Q.

i. Show that p+r=2q.

ii. Find in terms of a and q the equation of the locus of the midpoint M of Pr. State any restrictions on this locus.

I pretty much used parametrics with the mid point formula. When i get to y, it ends up being a(r^2+p^2)/2, which i complete the square for and then sub in the x parametric, but i am left with -2pr, which i can't eliminate. Also, what is the restriction of this locus?
 

dunjaaa

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????, the equation of chord PR: y=[(p+r)x]/2 - apr. Rearranging and substituting p+r with 2q using part (i), 2pr = [qx-2y]/a. Sub that in to find the locus. I'll let you find the restriction out.
 

braintic

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You don't need the y coordinate. (Nor do you need to find equations of chords)

The x-coordinate is x=a(p+r), which is 2aq from part (i), and that is a constant since Q is a fixed point.

Done. The equation is x=2aq.
 

aDimitri

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You don't need the y coordinate. (Nor do you need to find equations of chords)

The x-coordinate is x=a(p+r), which is 2aq from part (i), and that is a constant since Q is a fixed point.

Done. The equation is x=2aq.
this doesn't give the restriction

i just considered the fact that p =/= r and by looking at the way the midpoint moves it is quite clear that there is a range restriction of y > aq^2
 

dunjaaa

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Yeah, since Q is a fixed point the restriction is y>=aq^2 with equality iff p=q=r. If the locus is a vertical line, I guess you can deduce that restriction just based off a diagram.
 
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aDimitri

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Yeah, since Q is a fixed point the restriction is y>=aq^2 with equality iff p=q=r. If the locus is a vertical line, I guess you can deduce that restriction just based off a diagram.
no equality. we divided by p-r earlier in the question so p=/=r
 

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