parametric q. (1 Viewer)

[shkspeare]

hey just wondering how to do this question

a secant which passes through the point (0,-a) cuts the parabola x^2 = 4ay at P,Q with parametrs p,q
a)prove pq= 1
b)if S is the focus show that 1/PS + 1/QS = 1/a

 

[shkspeare]

actutaly nvm i got it alreadi haha

 

[shkspeare]

some1 do B for me plz im stuck >.<

 

[kazerati]

hey buds..

u kno, id really really love to be able to help, but im totally drawin a blank, which is NOT good...

so if ya get it, or if n e 1 gets it out for that matter, can i please know?

doh, im more skrewed than i thought...

 

[ND]

Originally posted by shkspeare
some1 do B for me plz im stuck >.<

All you would have to do is get the distances PS and QS (using pythag or whatever), which are obviouly both in the same form except for the paramater, then just sub into 1/PS + 1/QS, and it should come out quite easily. You'll have to use pq=-1 also, probably the last step before 1/a comes out.

 

[ND]

I'm bored so i'll do it anyway:

PS^2 = (2ap)^2 + (a(p^2-1))^2
= a^2(4p^2 + p^4 - 2p^2 + 1)
= a^2(p^2 + 1)^2
PS = a(p^2+1)
similarly, QS = a(q^2+1)

1/PS + 1/QS = (p^2+q^2+2)/a(p^2+1)(q^2+1)
= (p^2+q^2+2)/a((pq)^2+p^2+q^2+1)
= (p^2+q^2+2)/a(p^2+q^2+2)
= 1/a

 

[Rand]

HAHAHAHA, where'd ya get that Q from, it was in our school trial like 4 years ago...

 

[Affinity]

it's just one of the stock questions.

 

[shkspeare]

mm another question

if ur doing a geometric series, and ur using logarihtms to find out the nth term, are u meant to presume the ratio is positive (wen using logs)?

example:
which term of the series

8 - 4 + 2 - ... is equal to 1/128

a = 8, r = -0.5

1/128 = 8(-0.5)^n-1
(-0.5)^n-1 = 1/1024

now using logs u cant take the log of something negative...

i.e n = 1 + [ log(1/1024) / log(-1/2)]

i get the correct answer if i make the ratio positive

some1 clear this thing up plz? i'm pretty poor in this topic area and logs as well so yeah...

 

[iambored]

why can't you take the log of something negative?

 

[kimmeh]

Originally posted by iambored
why can't you take the log of something negative?

if you look at the log graph, it touches the x axis at one and it never touches the y axis [ie the y axis is an asymptote]


hmmm i got n = 11
if n is term number


Tn = ar^(n-1)
1/128 = 8 x (0.5)^(n-1)
1/1024 = (0.5)^(n-1)
(1024)^-1 = (0.5)^n x (0.5)^-1
(0.5)^n = 1/2048
.: n log (0.5) = log (1/2048)

.: n = 11


yeay ! it works XD

 

[:: ck ::]

n = 11

yesss same as kimmeh~! 4unit maths here i come

 

[kimmeh]

sure ryan.. u just copiend my answers

 

[:: ck ::]

bull! i dont need no 4unit maths girl to teach me how to do 3unit maths~

 

[shkspeare]

hey how would i find the tangent of y = x^2+1 @ 1,2
similarly with x^2-x-2=y @ (-2,4)

 

[kimmeh]

Originally posted by :: ryan.cck ::
bull! i dont need no 4unit maths girl to teach me how to do 3unit maths~

remeber ryan.. the books are STILL on my table



ummm


y = x^2 +1
dy/dx = 2x
.: m = 2x --> at (1,2)
sub x value into x
.: m = 2
--> using (y-y1) = m(x- x1)

(y - 2) = 2(x - 1)
.: y -2 = 2x - 2
.: y = 2x


for x^2 - x - 2 = y

dy/dx = 2x - 1
@ (-2, 4)
m = -5
--> using (y-y1) = m(x- x1)
(y- 4) = -5 (x - - 2)
.: (y-4) = -5x - 10
.: y = -5x + 6



simple ?

 

[kimmeh]

Originally posted by :: ryan.cck ::
bull! i dont need no 4unit maths girl to teach me how to do 3unit maths~

remeber ryan.. i have your books

Originally posted by shkspeare
hey how would i find the tangent of y = x^2+1 @ 1,2
similarly with x^2-x-2=y @ (-2,4)

y = x^2 +1
dy/dx = 2x
at (1,2)
m = 2(1)
.: m = 2
--> using (y - y1) = m (x - x1)
(y - 2) = 2 (x - 1)
y -2 = 2x - 2
.: y = 2x


for x^2 - x -2 = y

dy/dx = 2x -1
at (-2, 4)
m = 2(-2) -1
.: m = -5
--> using (y - y1) = m (x - x1)
(y - 4 ) = -5 ( x - -2 )
y - 4 = -5x - 10
.: y = -5x - 6


easy

 

[kimmeh]

whoops

 

[shkspeare]

is there a way of doing this using the tangent formula

 

[freaking_out]

Originally posted by shkspeare
is there a way of doing this using the tangent formula

i don't think so, coz ur dealing with a secant...but then again i could b wrong.