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If the chord of contact of the tangents to a parabola x^2=4ay, from an external point T(x1,y1) meets the directrix at R, prove that RT subtends a right angle at the focus.

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mmmmm, you werent kidding around
Is this easy for you ? xD
i will have to think about it

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Other answers:

the directrix is that line under the parabola
lols, you were thinking that its easy ? xD
Yup, y=-a
what class is this ?
I live in Australia, so its kinda different, there's only a maths class xD
sorry wrong question, have you done calculus
Yes I have, how did you know ? xD
What grade are you in ?
it will definitely help
im a sophomore in college
Cool~ college, and you havent learnt that ?
not specifically, but i bet i can do this
lols, then try hehe
im trying to make a diagram, im not even sure i understand the question yet
ok you have an external pt, and you want the tangent through it
Yeah, I don't really know where's the external point is, thats why im lost =/
ok first thing is first. the chord of contact means something special
Something special ? What do you mean ?
there is a picture of a chord of contacts http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf
well given an external point there are two tangents from that point to the parabola
now where the tangents intersect the parabola, now draw a chord through those two points
where the tangents meet the parabola, the two tangents
wow~ nice pdf file, its detailed on parametrics ty
But where is the external point ?
anywhere
Can't it have to be specific
in the pdf, the point is below the parabola
It can be but it doesn't make sense when it says RT subtends a right angle at the focus.
like this
like this ?
|dw:1319452686670:dw|
I know but where's the external point ?
|dw:1319452750463:dw|
Imaginary ?
|dw:1319452790390:dw|
Is the parabola imaginery or somethinbg ?
imaginary* something*
|dw:1319452894222:dw|
I don't get your diagrams ? sorry
|dw:1319453015665:dw|
I don't get it ? >_<
yeah the diagram is a beast
lols, why is there like 3 digrams ? i dont get it >_<
|dw:1319453240562:dw|
lols, this is not 3D trig xD
can you do better ?
Nope, i dont know how to start, the problem is where is T
just a point below the parabola
like in that pdf
But it won't make sense, and i dont know how to prove
have faith
have faith ?
i think they mean the segment R T
yes have faith when you draw it that it will all come into place, the problem will unravel
i am atheist, what did you think i mean ?
idk, never mind then i will work it out myself
|dw:1319453605908:dw|
ty (:
im working on it
would be nice if they gave you a diagram
Yeah i know, none of the hard questions have diagrams =/
i think RT is the line segment though
hm idk
|dw:1319454238070:dw|
\[\text{An equation of the parabola with focus }(0,p)\text{ and directrix }y = -p\text{ is}\] \[x^2=4py\]
so the directrix would be y = -a in your question :-D
yes
Yeah, but im having problem proving
If the chord of contact of the tangents to a parabola x² = 4ay, from an external point T(x1, y1) meets the directrix at R, prove that RT subtends a right angle at the focus. chord of contact of tangents to parabola is xx1=2a(y+y1) directrix is y=-a substitute into that, x=[2a(-a+y1)]/x1 y=-a R([2a(-a+y1)]/x1,-a) let S be the focus d/dx RS = 2a/ -[2a(-a+y1)]/x1= x1/(a-y1) d/dx TS=(y1-a)/x1 d/dx RS * d/dx TS= -1
lols, you got it from boredofstudies right ?
yes... :-P
how did you get chord of tangents equation
hey, can you draw a diagram then of whats going on, please post the link
Yeah but the problem is that I don't understand the solution
it helps to draw a diagram.
can you post the link to the solution please , so i can read it
http://community.boredofstudies.org/showthread.php?t=192825&page=1
But what's xx1 ? im trying to figure it out
http://wiki.answers.com/Q/Find_the_equation_of_chord_of_contact_tangent_to_the_parabola
oh i get it, but where did y=1/2(p+q)x-apq come from ?
xx1 must mean something like \[x_1\times x= 2a(y+y_1)\]
where did 2a come from ?
good question
\[parabola: x^2 = 4py\]\[a = \frac{1}{4p}\]
no, isn't it x^2 = 4ay y = x^2/4a y' = 2x/4a = x/2a
|dw:1319455562515:dw|
i wish there were other colors, but from T draw two tangents to the parabola
perl's diagram works.... but how do we get there using calculus and derivatives if we don't even know the equation for chord of contact of the tangents to a parabola
oh the equation of the tangent to the parabola is y=px-ap^2 right ?
|dw:1319455702848:dw|
they are given here http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf
the chords of tangents formula
now it all makes sense
|dw:1319455885258:dw|
i dont get that parametrisation, is it because they want x^2 = 4ay ? , what is p
Doesn't it involve parameters as well P(2ap,ap^2)
they want something like the quotient of the slopes between 2 lines is equal to -1
and one of those lines should be RT
|dw:1319456032235:dw|
crap, it missed the R it is the right bottom corner
Im confused, is parameters involved ?
no parameter seems to be involved :-(
actually a is the parameter, you have to be careful, this word parameter is used in different senses
, for instance in y = mx + b , m and b are parameters . now another sense of parameter is used when we 'parametrize' an equation. this means we introduce a third variable, often t
isn't parameters P(2ap,ap^2) So there must be a point for the parameters ?
meh, I assumed m and b to be constant
right, but they are considered 'parameters'
but its not a straight line so it cant be y=mx + b
in the sense that they change your line . so for x^2 = 4ay, the parameter is a since that produces different parabolas by tweaking a
think of in this sense a parameter is your tweaking constant
:-(
the parameter is 'a'*
I thought a was just a slope
Im lost =/ Parameter P(2ap,ap^2)_
changing 'a' changes your parabola. but for the line y = mx + b, changing m,b changes your line
these are your tweaking values .
I dont like their use of p, scratch that and use t instead
ok they are using 'parametric equations' , x = f(t), y = g(t)
so x = 2at, y = at^2
I like to use matrices and vectors.
Yeah thats in cartesian form
ok, use a matrix or vector, id like to see that
mimi, when you remove t, you get cartesian equation
matrix or vector ?
Oh yeah my bad i forgot xD
http://img392.imageshack.us/img392/8476/lach11n72fb.jpg
I haven't learnt that complicated stuff yet =/
you will learn it soon.
Hope not looks hard already =/
thats curve fitting, no big deal
you will find it easier than conics
Whats conics ?
its a quick way to fit a quadratic sequence, cubic, etc
mimi, what we are doing is conics, oh dea
conic sections: parabolas, circles, ellipses, hyperbolas, stuff.
oh dear *
you do them along parametric equations usually.
agdg, put the proper definition on board ;)
well, they're what you get when you cut a cone.
LOL, this is not conics
a conic has a specific algebraic equation in x and y , quadratic Ax^2 + Bxy + cY^2 + dy + ex + f = 0
yes it is, a parabola is a conic
a conic is a family of curves, as agdg pointed out
But Ax^2 + Bxy + cY^2 + dy + ex + f = 0 ?
forget the algebra and xs and ys and f(x)s, you will understand them really well once you picture them geometrically
http://mathworld.wolfram.com/ConicSection.html
right
Um, this doesnt look like a conic
agdg, whats the nth term of this quadratic sequence 3,5,10,18...
lets test your matrix skills
hmmm... is it infinity?
no, the nth term
give me a formula for the sequence
obviously it diverges, ;)
its \[2^n + 2^\frac{n}{2}\]
no, lol
wait! it's \[\frac{2^n+2^{\frac{n}{2}}}{2}\]
its a quadratic sequence, like in your matrix
how do we find it :-D
so what's the 5th term
ok so first we find the differences
the first row of differences is 2 , 5, 8, 11 , ... the second row of differences is 3, 3, 3, ... so you know its a quadratic sequence
that looks like a long matrix
3, 5, 10, 18, 29 ... thats the sequence
it should fit f(n) = an^2 + bn + c
I guess that works.
how do we multiply matrices?
are you asking in general ?

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