Parametric equations of Parabola - Need help (1 Viewer)

[blackops23]

Hi guys, here's the question:

Q. A secant passes through (0,-a) and cuts the parabola x^2 =4ay at P,Q with respective parameters p,q. pq = 1.

If S is the focus, show that 1/PS + 1/QS = 1/a
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Anyway to do this without using the distance formula??
Thanks guys

 

[Bored Of Fail 2]

why would you want to do it any other way??

using distance formula you get PS = a (p^2 +1 )

similarly QS = a ( q^2 +1 )

then LHS = (1/a) [ 1/ (q^2 +1 ) + 1/ (p^2 +1 ) ]
then get common denomintor


LHS = (1/a) [ ( q^2 +1 +p^2 +1 ) / ( p^2q^2 +p^2 +q^2 +1 ) ]

by expanding the bottom

now use the result p^2 +q^2 = (p+q)^2 -2pq and it all falls out

 

[Bored Of Fail 2]

LHS = (1/a) [ (p+q)^2 -2pq +2 ] / ( (pq)^2 +(p+q)^2 -2pq +1 )

then sub pq=1, and cancel down

 

[deterministic]

Easier way than distance formula:

Remember the fundamental fact of the parabola:
distance from any point on the parabola to the focus = perpendicular distance to the directrix. Noting directrix is y=-a, so:

PS = distance to focus = perp. distance from P to directrix = ap^2+a (draw a diagram to convince yourself)

Similar for QS and just follow Bored for the algebra.

 

[Drongoski]

deterministic

You are a young genius; should be doing extremely well in yr HSC.

 

[Bored Of Fail 2]

deterministic

You are a young genius; should be doing extremely well in yr HSC.

:S, where have you been , he did his hsc in 2009 im pretty sure. Also he did 4unit math for hsc and is doing advanced maths at uni

 

[slyhunter]

Still a nice compliment though. =)

 

[Drongoski]

:S, where have you been , he did his hsc in 2009 im pretty sure. Also he did 4unit math for hsc and is doing advanced maths at uni

Bored of Fail 2

Thanks for pointing out my mistake. I thought he was doing Yr11 or something. Anyway he is damn good.

 

[deterministic]

Cheers Drongoski =D

Bored is right though, I did my HSC in 09 but im doing maths and finance, not adv maths. I had another account in 2009 but I didnt use it much in HSC so I decided to start fresh. I just enjoy maths so I come here to help others when needed.

 

[blackops23]

why would you want to do it any other way??

using distance formula you get PS = a (p^2 +1 )

similarly QS = a ( q^2 +1 )

then LHS = (1/a) [ 1/ (q^2 +1 ) + 1/ (p^2 +1 ) ]
then get common denomintor


LHS = (1/a) [ ( q^2 +1 +p^2 +1 ) / ( p^2q^2 +p^2 +q^2 +1 ) ]

by expanding the bottom

now use the result p^2 +q^2 = (p+q)^2 -2pq and it all falls out

I must be missing something here or maybe i'm just a little slow...

P=(2ap, ap^2)
S= (0,a)

PS=sqrt[(2ap)^2 + (a^2)(p^2 + 1)^2)]
=a*sqrt(4p^2 + (p^2+1)^2)

How do I get PS= a(p^2 + 1)? i.e. how does 2ap = 0?

Oh yeah, and thanks deterministic, bored of fail 2, and any other guys that help on BOS, you're all champs =)

 

[Bored Of Fail 2]

P=(2ap, ap^2)
S= (0,a)

PS = sqrt ( (2ap-0)^2 + ( ap^2 -a)^2 )
= sqrt ( 4a^2p^2 + a^2p^4 -2a^2p^2 +a^2 )
= sqrt ( a^2 [ p^4 +2a^2p^2 +1] ) = sqrt [ a^2 (p^2+1)^2) = a (p^2 +1)

that second factor under sqrt is a perfect square

 

[Bored Of Fail 2]

I must be missing something here or maybe i'm just a little slow...

P=(2ap, ap^2)
S= (0,a)

PS=sqrt[(2ap)^2 + (a^2)(p^2 + 1)^2)] // that plus should be a minus as well in the second bracket
=a*sqrt(4p^2 + (p^2+1)^2)

How do I get PS= a(p^2 + 1)? i.e. how does 2ap = 0?

Oh yeah, and thanks deterministic, bored of fail 2, and any other guys that help on BOS, you're all champs =)

see // comment above

 

[blackops23]

Easier way than distance formula:

Remember the fundamental fact of the parabola:
distance from any point on the parabola to the focus = perpendicular distance to the directrix. Noting directrix is y=-a, so:

PS = distance to focus = perp. distance from P to directrix = ap^2+a (draw a diagram to convince yourself)

Similar for QS and just follow Bored for the algebra.

pretty damn nice... =)

Thanks mate