Question on locus!! NEED HELP! - Cambridge (1 Viewer)

[slix_88]

Hi,

I can't do this question from Cambridge YR11 CH9:EX 9E Q7c)

Q7 P and Q are the points with parameters p and q on the parabola x = 2at and y = at^2.
A) Show that the chord PQ is y - 1/2(p+q)x + apq = 0 - DONE THAT
B) IF the chord when extended passes through the point (0, -a), show that pq = 1 - DONE THAT
C) Hence, if S is the focus of the parabola, show that 1/SP + 1/SQ = 1/a ???????

How do you do C?

 

[香港!]

Hi,

I can't do this question from Cambridge YR11 CH9:EX 9E Q7c)

Q7 P and Q are the points with parameters p and q on the parabola x = 2at and y = at^2.
A) Show that the chord PQ is y - 1/2(p+q)x + apq = 0 - DONE THAT
B) IF the chord when extended passes through the point (0, -a), show that pq = 1 - DONE THAT
C) Hence, if S is the focus of the parabola, show that 1/SP + 1/SQ = 1/a ???????

How do you do C?

the cuve is x²=4ay obtained from x=2at, y=at²
so focus is S=(0,a)
Distance
(SP)²=(2ap-0)²+(ap²-a)²
=4a²p²+a²p^4-2a²p²+a²
=2a²p²+a²p^4+a²
=a²(p^4+2p²+1)
=a²(p²+1)²
.: SP=a(p²+1)
similarly, SQ=a(q²+1)


1\SP+1\SQ
=1\a(p²+1)+1\a(q²+1)
=[(q²+1)+(p²+1)]\a(q²+1)(p²+1)
=(p²+q²+2)\a(q²+1)(p²+1)

but u have shown that pq=1
so q=1\p
.: 1\SP+1\SQ=(p²+1\p²+2)\a(1\p²+1)(p²+1)
times p² on top and bottom
1\SP+1\SQ=(p^4+2p²+1)\a(1+p²)²
=(p²+1)²\a(1+p²)²
=1\a

 

[slix_88]

Thanx a lot...

I also have 2 more questions...

9. Using the equation y = 1/2(p+q)x - apq, show that the midpoint of the chord of the parabola x = 2at, y = at^2 lies on the vertical line x = k if and only if the chord has a gradient k/2a.

10. The points P, Q, R and S lie on x = 2at, y = at^2 and have paramenters p, q, r and s respectively. If the chords PQ and RS intersect on the axis, show that p:r = s:q.

THANX A LOT!

I PROMISE THESE ARE THE LAST ONES!

 

[shafqat]

For the question above, there's another way to find SP if you want. Drop perpendicular down to directrix, let the line be PM. PM = ap^2 + a = PS (definition of parabola)

 

[slix_88]

Can you do question 9 and 10? If so, could u?

Should I worry a lot about these type of questions? Im in year 11 now. Would this be further covered in the YR 12 3U and 4U course???

 

[香港!]

Thanx a lot...

I also have 2 more questions...

9. Using the equation y = 1/2(p+q)x - apq, show that the midpoint of the chord of the parabola x = 2at, y = at^2 lies on the vertical line x = k if and only if the chord has a gradient k/2a.

10. The points P, Q, R and S lie on x = 2at, y = at^2 and have paramenters p, q, r and s respectively. If the chords PQ and RS intersect on the axis, show that p:r = s:q.

THANX A LOT!

I PROMISE THESE ARE THE LAST ONES!

HiHi, u dun need to "PROMISE THESE ARE THE LAST ONES!"
feel free to ask as many q's as u want!!

9.midpoint of the chord is
x=(2ap+2aq)\2
=a(p+q)
y=(ap²+aq²)\2
=a(p²+q²)\2

from the given equation, gradient of the chord is
(p+q)\2
so "if and only if the chord has a gradient k/2a"
then (p+q)\2=k\2a
a(p+q)=k
we've shown x=a(p+q) in the midpoint of the chord PQ
so x=k

10.
eq. chord PQ: y = 1/2(p+q)x - apq
eq. chord RS: y = 1/2(r+s)x - ars

"If the chords PQ and RS intersect on the axis"
then it's either at x=0 or y=0
when x=0, y=-apq and y=-ars
so -apq=-ars
pq=rs
p\r=s\q
.: p:r=s:q

 

[slix_88]

WOAH!

Are you a math teacher? Far out you know so much!!! When are you/have done your HSC? Are you doing/have done 4U?? is that why u know so much...

THNX FOR ALL THE HELP!

PS: PM me your email - THNX

 

[who_loves_maths]

^ lol ... nah, he's only doing his HSC this year {i know... incredulous isn't it. :uhhuh:}, and yes he's doing 4u. but no, he's not a maths teacher, which makes it even more amazing that he's probably better than most of them

 

[香港!]

^lol who_loves_maths, ur da one who truely knows sooo much!
not me, the questions i do are simple, u juz need practice den u can do dem easily
but the questions who_loves_maths does is really AMAZING
he is really the amazing one!!

 

[who_loves_maths]

^ but simplicity is what maths should be about ... and practice is what makes maths seem simple

so dont' listen to him slix_88 , he's just jesting around .

 

[香港!]

slix_88 go to the extension 2 forum and take a look of some of his posts...
you will be amazed for sure
many other people said "wow" after his posts if you look carefully ^_^

 

[Dumsum]

I say you're both geniuses.

 

[who_loves_maths]

Originally Posted by 香港!
slix_88 go to the extension 2 forum and take a look of some of his posts...
you will be amazed for sure
many other people said "wow" after his posts if you look carefully ^_^

haha... exaggeration! ... one person doesn't = many ppl
and yes you do have to squeeze your eyes and look carefully before you can see it .

oh and slix_88, he forgot to mention that it's only because of the length of my posts {where most ppl don't even bother finish reading 'em}, and not the quality of them!

if you want to see true mathematical elegance, go to the 4u forum and look at 香港!'s (alias FinalFantasy) solutions to problems - they're always so short and compact... yet very comprehensive !

 

[mojako]

香港!'s (alias FinalFantasy)

he/she's not FinalFantasy lol

 

[香港!]

^ yap that's... right!!!

 

[who_loves_maths]

Originally Posted by mojako
he/she's not FinalFantasy lol

right ... must be my fault then ... lol .

hehe ... then i wonder where FinalFantasy went ...

 

[mojako]

right ... must be my fault then ... lol .

hehe ... then i wonder where FinalFantasy went ...

he went to HK
so its unlikely that he'll be answering maths q from over there
unless he loves maths more than u do

 

[who_loves_maths]

Originally Posted by mojako
he went to HK
so its unlikely that he'll be answering maths q from over there

hehe... yea, i'd take a holiday just before my HSC as well... except i'm too busy these day - you know, being occupied by the HSC and all...

Originally Posted by mojako
unless he loves maths more than u do

lol ...

 

[sikeveo]

??! sounds like FF. Being so damn asian and all.

 

[香港!]

nah man, FF told me he went on holidays