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.

- Mimi_x3

- katieb

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- perl

mmmmm, you werent kidding around

- Mimi_x3

Is this easy for you ? xD

- perl

i will have to think about it

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## More answers

- perl

the directrix is that line under the parabola

- Mimi_x3

lols, you were thinking that its easy ? xD

- Mimi_x3

Yup, y=-a

- perl

what class is this ?

- Mimi_x3

I live in Australia, so its kinda different, there's only a maths class xD

- perl

sorry wrong question, have you done calculus

- Mimi_x3

Yes I have, how did you know ? xD

- Mimi_x3

What grade are you in ?

- perl

it will definitely help

- perl

im a sophomore in college

- Mimi_x3

Cool~ college, and you havent learnt that ?

- perl

not specifically, but i bet i can do this

- Mimi_x3

lols, then try hehe

- perl

im trying to make a diagram, im not even sure i understand the question yet

- perl

ok you have an external pt, and you want the tangent through it

- Mimi_x3

Yeah, I don't really know where's the external point is, thats why im lost =/

- perl

ok first thing is first. the chord of contact means something special

- Mimi_x3

Something special ? What do you mean ?

- perl

there is a picture of a chord of contacts http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

- perl

well given an external point there are two tangents from that point to the parabola

- perl

now where the tangents intersect the parabola, now draw a chord through those two points

- perl

where the tangents meet the parabola, the two tangents

- Mimi_x3

wow~ nice pdf file, its detailed on parametrics ty

- Mimi_x3

But where is the external point ?

- perl

anywhere

- Mimi_x3

Can't it have to be specific

- perl

in the pdf, the point is below the parabola

- Mimi_x3

It can be but it doesn't make sense when it says RT subtends a right angle at the focus.

- perl

like this

- Mimi_x3

like this ?

- perl

|dw:1319452686670:dw|

- Mimi_x3

I know but where's the external point ?

- perl

|dw:1319452750463:dw|

- Mimi_x3

Imaginary ?

- perl

|dw:1319452790390:dw|

- Mimi_x3

Is the parabola imaginery or somethinbg ?

- Mimi_x3

imaginary* something*

- perl

|dw:1319452894222:dw|

- Mimi_x3

I don't get your diagrams ? sorry

- perl

|dw:1319453015665:dw|

- Mimi_x3

I don't get it ? >_<

- perl

yeah the diagram is a beast

- Mimi_x3

lols, why is there like 3 digrams ? i dont get it >_<

- perl

|dw:1319453240562:dw|

- Mimi_x3

lols, this is not 3D trig xD

- perl

can you do better ?

- Mimi_x3

Nope, i dont know how to start, the problem is where is T

- perl

just a point below the parabola

- perl

like in that pdf

- Mimi_x3

But it won't make sense, and i dont know how to prove

- perl

have faith

- Mimi_x3

have faith ?

- perl

i think they mean the segment R T

- perl

yes have faith when you draw it that it will all come into place, the problem will unravel

- perl

i am atheist, what did you think i mean ?

- Mimi_x3

idk, never mind then i will work it out myself

- perl

|dw:1319453605908:dw|

- Mimi_x3

ty (:

- perl

im working on it

- perl

would be nice if they gave you a diagram

- Mimi_x3

Yeah i know, none of the hard questions have diagrams =/

- perl

i think RT is the line segment though

- Mimi_x3

hm idk

- perl

|dw:1319454238070:dw|

- anonymous

\[\text{An equation of the parabola with focus }(0,p)\text{ and directrix }y = -p\text{ is}\]
\[x^2=4py\]

- anonymous

so the directrix would be y = -a in your question :-D

- perl

yes

- Mimi_x3

Yeah, but im having problem proving

- anonymous

If the chord of contact of the tangents to a parabola xÂ² = 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

- Mimi_x3

lols, you got it from boredofstudies right ?

- anonymous

yes... :-P

- perl

how did you get chord of tangents equation

- perl

hey, can you draw a diagram then of whats going on, please post the link

- Mimi_x3

Yeah but the problem is that I don't understand the solution

- anonymous

it helps to draw a diagram.

- perl

can you post the link to the solution please , so i can read it

- anonymous

http://community.boredofstudies.org/showthread.php?t=192825&page=1

- Mimi_x3

But what's xx1 ? im trying to figure it out

- anonymous

http://wiki.answers.com/Q/Find_the_equation_of_chord_of_contact_tangent_to_the_parabola

- Mimi_x3

oh i get it, but where did y=1/2(p+q)x-apq come from ?

- anonymous

xx1 must mean something like \[x_1\times x= 2a(y+y_1)\]

- Mimi_x3

where did 2a come from ?

- anonymous

good question

- anonymous

\[parabola: x^2 = 4py\]\[a = \frac{1}{4p}\]

- Mimi_x3

no, isn't it x^2 = 4ay
y = x^2/4a
y' = 2x/4a = x/2a

- perl

|dw:1319455562515:dw|

- perl

i wish there were other colors, but from T draw two tangents to the parabola

- anonymous

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

- Mimi_x3

oh the equation of the tangent to the parabola is y=px-ap^2 right ?

- perl

|dw:1319455702848:dw|

- perl

they are given here http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

- perl

the chords of tangents formula

- anonymous

now it all makes sense

- perl

|dw:1319455885258:dw|

- perl

i dont get that parametrisation, is it because they want x^2 = 4ay ? , what is p

- Mimi_x3

Doesn't it involve parameters as well P(2ap,ap^2)

- anonymous

they want something like the quotient of the slopes between 2 lines is equal to -1

- anonymous

and one of those lines should be RT

- perl

|dw:1319456032235:dw|

- perl

crap, it missed the R it is the right bottom corner

- Mimi_x3

Im confused, is parameters involved ?

- anonymous

no parameter seems to be involved :-(

- perl

actually a is the parameter, you have to be careful, this word parameter is used in different senses

- perl

, 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

- Mimi_x3

isn't parameters P(2ap,ap^2)
So there must be a point for the parameters ?

- anonymous

meh, I assumed m and b to be constant

- perl

right, but they are considered 'parameters'

- Mimi_x3

but its not a straight line so it cant be y=mx + b

- perl

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

- perl

think of in this sense a parameter is your tweaking constant

- anonymous

:-(

- perl

the parameter is 'a'*

- anonymous

I thought a was just a slope

- Mimi_x3

Im lost =/ Parameter P(2ap,ap^2)_

- perl

changing 'a' changes your parabola. but for the line y = mx + b, changing m,b changes your line

- perl

these are your tweaking values .

- perl

I dont like their use of p, scratch that and use t instead

- perl

ok they are using 'parametric equations' , x = f(t), y = g(t)

- perl

so x = 2at, y = at^2

- anonymous

I like to use matrices and vectors.

- Mimi_x3

Yeah thats in cartesian form

- perl

ok, use a matrix or vector, id like to see that

- perl

mimi, when you remove t, you get cartesian equation

- Mimi_x3

matrix or vector ?

- Mimi_x3

Oh yeah my bad i forgot xD

- anonymous

http://img392.imageshack.us/img392/8476/lach11n72fb.jpg

- Mimi_x3

I haven't learnt that complicated stuff yet =/

- anonymous

you will learn it soon.

- Mimi_x3

Hope not looks hard already =/

- perl

thats curve fitting, no big deal

- anonymous

you will find it easier than conics

- Mimi_x3

Whats conics ?

- perl

its a quick way to fit a quadratic sequence, cubic, etc

- perl

mimi, what we are doing is conics, oh dea

- anonymous

conic sections: parabolas, circles, ellipses, hyperbolas, stuff.

- perl

oh dear *

- anonymous

you do them along parametric equations usually.

- perl

agdg, put the proper definition on board ;)

- anonymous

well, they're what you get when you cut a cone.

- Mimi_x3

LOL, this is not conics

- perl

a conic has a specific algebraic equation in x and y , quadratic
Ax^2 + Bxy + cY^2 + dy + ex + f = 0

- perl

yes it is, a parabola is a conic

- perl

a conic is a family of curves, as agdg pointed out

- Mimi_x3

But Ax^2 + Bxy + cY^2 + dy + ex + f = 0 ?

- anonymous

forget the algebra and xs and ys and f(x)s, you will understand them really well once you picture them geometrically

- anonymous

http://mathworld.wolfram.com/ConicSection.html

- perl

right

- Mimi_x3

Um, this doesnt look like a conic

- perl

agdg, whats the nth term of this quadratic sequence
3,5,10,18...

- perl

lets test your matrix skills

- anonymous

hmmm... is it infinity?

- perl

no, the nth term

- perl

give me a formula for the sequence

- perl

obviously it diverges, ;)

- anonymous

its \[2^n + 2^\frac{n}{2}\]

- perl

no, lol

- anonymous

wait! it's \[\frac{2^n+2^{\frac{n}{2}}}{2}\]

- perl

its a quadratic sequence, like in your matrix

- anonymous

how do we find it :-D

- anonymous

so what's the 5th term

- perl

ok so first we find the differences

- perl

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

- anonymous

that looks like a long matrix

- perl

3, 5, 10, 18, 29 ... thats the sequence

- perl

it should fit f(n) = an^2 + bn + c

- anonymous

I guess that works.

- anonymous

how do we multiply matrices?

- perl

are you asking in general ?

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