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Parametrics question (1 Viewer)

elseany

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I cant seem to get this one out:

P(2ap,ap2) & Q(2aq,aq2) lie on the parabola x2 = 4ay, where a > 0. The chord PQ passes through the focus.
Show that the chord PQ has length A(p + 1/p)2

thanks for your help
 

elseany

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yeh i can find everything except it doesnt seem to work out when i do it with the distance formula.

i've already shown that pq = -1

but my working gets really messy with the distance formula and just doesnt solve :<
 

jyu

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elseany said:
yup lol, i remembered that.

can somebody give the algebra a go?
Length = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
=sqrt[(2ap -2aq)^2 + (ap^2 - aq^2)^2]
=sqrt[4a^2(p - q)^2 + a^2(p^2 - q^2)^2]
=sqrt[4a^2(p + 1/p)^2 + a^2(p^2 - 1/p^2)^2]
=a sqrt[4(p + 1/p)^2 + (p^2 - 1/p^2)^2]
=a sqrt[(p + 1/p)^2 (4 + (p - 1/p)^2)]
=a sqrt[(p + 1/p)^2 (p + 1/p)^2]
=a(p + 1/p)^2

:santa: :santa: :santa:
 

Affinity

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Or:

let S be the focus (0,a), we know that SP=ap^2 + a = a(p^2 + 1)
SQ = aq^2 + a = a(1/p^2 + 1) [think why?]
so PQ = QS + SP = a (p^2 + 2 + 1/p^2) = a(p + 1/p)^2
 

elseany

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thanks jyu and affinity, but theres something i dont get;


=a sqrt[(p + 1/p)^2 (4 + (p - 1/p)^2)]
=a sqrt[(p + 1/p)^2 (p + 1/p)^2]

how do you go from that first line to that second line?
 
P

pLuvia

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Try expanding (4 + (p - 1/p)^2)] and it should turn out to be the same as (p + 1/p)^2]
 

Nktnet

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Or:

let S be the focus (0,a), we know that SP=ap^2 + a = a(p^2 + 1)
SQ = aq^2 + a = a(1/p^2 + 1) [think why?]
so PQ = QS + SP = a (p^2 + 2 + 1/p^2) = a(p + 1/p)^2
Thanks this helps, (12 years later)!
 

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