Mimi_x3
  • Mimi_x3
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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perl
  • perl
mmmmm, you werent kidding around
Mimi_x3
  • Mimi_x3
Is this easy for you ? xD
perl
  • perl
i will have to think about it

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perl
  • perl
the directrix is that line under the parabola
Mimi_x3
  • Mimi_x3
lols, you were thinking that its easy ? xD
Mimi_x3
  • Mimi_x3
Yup, y=-a
perl
  • perl
what class is this ?
Mimi_x3
  • Mimi_x3
I live in Australia, so its kinda different, there's only a maths class xD
perl
  • perl
sorry wrong question, have you done calculus
Mimi_x3
  • Mimi_x3
Yes I have, how did you know ? xD
Mimi_x3
  • Mimi_x3
What grade are you in ?
perl
  • perl
it will definitely help
perl
  • perl
im a sophomore in college
Mimi_x3
  • Mimi_x3
Cool~ college, and you havent learnt that ?
perl
  • perl
not specifically, but i bet i can do this
Mimi_x3
  • Mimi_x3
lols, then try hehe
perl
  • perl
im trying to make a diagram, im not even sure i understand the question yet
perl
  • perl
ok you have an external pt, and you want the tangent through it
Mimi_x3
  • Mimi_x3
Yeah, I don't really know where's the external point is, thats why im lost =/
perl
  • perl
ok first thing is first. the chord of contact means something special
Mimi_x3
  • Mimi_x3
Something special ? What do you mean ?
perl
  • perl
there is a picture of a chord of contacts http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf
perl
  • perl
well given an external point there are two tangents from that point to the parabola
perl
  • perl
now where the tangents intersect the parabola, now draw a chord through those two points
perl
  • perl
where the tangents meet the parabola, the two tangents
Mimi_x3
  • Mimi_x3
wow~ nice pdf file, its detailed on parametrics ty
Mimi_x3
  • Mimi_x3
But where is the external point ?
perl
  • perl
anywhere
Mimi_x3
  • Mimi_x3
Can't it have to be specific
perl
  • perl
in the pdf, the point is below the parabola
Mimi_x3
  • Mimi_x3
It can be but it doesn't make sense when it says RT subtends a right angle at the focus.
perl
  • perl
like this
Mimi_x3
  • Mimi_x3
like this ?
perl
  • perl
|dw:1319452686670:dw|
Mimi_x3
  • Mimi_x3
I know but where's the external point ?
perl
  • perl
|dw:1319452750463:dw|
Mimi_x3
  • Mimi_x3
Imaginary ?
perl
  • perl
|dw:1319452790390:dw|
Mimi_x3
  • Mimi_x3
Is the parabola imaginery or somethinbg ?
Mimi_x3
  • Mimi_x3
imaginary* something*
perl
  • perl
|dw:1319452894222:dw|
Mimi_x3
  • Mimi_x3
I don't get your diagrams ? sorry
perl
  • perl
|dw:1319453015665:dw|
Mimi_x3
  • Mimi_x3
I don't get it ? >_<
perl
  • perl
yeah the diagram is a beast
Mimi_x3
  • Mimi_x3
lols, why is there like 3 digrams ? i dont get it >_<
perl
  • perl
|dw:1319453240562:dw|
Mimi_x3
  • Mimi_x3
lols, this is not 3D trig xD
perl
  • perl
can you do better ?
Mimi_x3
  • Mimi_x3
Nope, i dont know how to start, the problem is where is T
perl
  • perl
just a point below the parabola
perl
  • perl
like in that pdf
Mimi_x3
  • Mimi_x3
But it won't make sense, and i dont know how to prove
perl
  • perl
have faith
Mimi_x3
  • Mimi_x3
have faith ?
perl
  • perl
i think they mean the segment R T
perl
  • perl
yes have faith when you draw it that it will all come into place, the problem will unravel
perl
  • perl
i am atheist, what did you think i mean ?
Mimi_x3
  • Mimi_x3
idk, never mind then i will work it out myself
perl
  • perl
|dw:1319453605908:dw|
Mimi_x3
  • Mimi_x3
ty (:
perl
  • perl
im working on it
perl
  • perl
would be nice if they gave you a diagram
Mimi_x3
  • Mimi_x3
Yeah i know, none of the hard questions have diagrams =/
perl
  • perl
i think RT is the line segment though
Mimi_x3
  • Mimi_x3
hm idk
perl
  • perl
|dw:1319454238070:dw|
anonymous
  • anonymous
\[\text{An equation of the parabola with focus }(0,p)\text{ and directrix }y = -p\text{ is}\] \[x^2=4py\]
anonymous
  • anonymous
so the directrix would be y = -a in your question :-D
perl
  • perl
yes
Mimi_x3
  • Mimi_x3
Yeah, but im having problem proving
anonymous
  • anonymous
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
Mimi_x3
  • Mimi_x3
lols, you got it from boredofstudies right ?
anonymous
  • anonymous
yes... :-P
perl
  • perl
how did you get chord of tangents equation
perl
  • perl
hey, can you draw a diagram then of whats going on, please post the link
Mimi_x3
  • Mimi_x3
Yeah but the problem is that I don't understand the solution
anonymous
  • anonymous
it helps to draw a diagram.
perl
  • perl
can you post the link to the solution please , so i can read it
anonymous
  • anonymous
http://community.boredofstudies.org/showthread.php?t=192825&page=1
Mimi_x3
  • Mimi_x3
But what's xx1 ? im trying to figure it out
anonymous
  • anonymous
http://wiki.answers.com/Q/Find_the_equation_of_chord_of_contact_tangent_to_the_parabola
Mimi_x3
  • Mimi_x3
oh i get it, but where did y=1/2(p+q)x-apq come from ?
anonymous
  • anonymous
xx1 must mean something like \[x_1\times x= 2a(y+y_1)\]
Mimi_x3
  • Mimi_x3
where did 2a come from ?
anonymous
  • anonymous
good question
anonymous
  • anonymous
\[parabola: x^2 = 4py\]\[a = \frac{1}{4p}\]
Mimi_x3
  • Mimi_x3
no, isn't it x^2 = 4ay y = x^2/4a y' = 2x/4a = x/2a
perl
  • perl
|dw:1319455562515:dw|
perl
  • perl
i wish there were other colors, but from T draw two tangents to the parabola
anonymous
  • 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
  • Mimi_x3
oh the equation of the tangent to the parabola is y=px-ap^2 right ?
perl
  • perl
|dw:1319455702848:dw|
perl
  • perl
they are given here http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf
perl
  • perl
the chords of tangents formula
anonymous
  • anonymous
now it all makes sense
perl
  • perl
|dw:1319455885258:dw|
perl
  • perl
i dont get that parametrisation, is it because they want x^2 = 4ay ? , what is p
Mimi_x3
  • Mimi_x3
Doesn't it involve parameters as well P(2ap,ap^2)
anonymous
  • anonymous
they want something like the quotient of the slopes between 2 lines is equal to -1
anonymous
  • anonymous
and one of those lines should be RT
perl
  • perl
|dw:1319456032235:dw|
perl
  • perl
crap, it missed the R it is the right bottom corner
Mimi_x3
  • Mimi_x3
Im confused, is parameters involved ?
anonymous
  • anonymous
no parameter seems to be involved :-(
perl
  • perl
actually a is the parameter, you have to be careful, this word parameter is used in different senses
perl
  • 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
  • Mimi_x3
isn't parameters P(2ap,ap^2) So there must be a point for the parameters ?
anonymous
  • anonymous
meh, I assumed m and b to be constant
perl
  • perl
right, but they are considered 'parameters'
Mimi_x3
  • Mimi_x3
but its not a straight line so it cant be y=mx + b
perl
  • 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
  • perl
think of in this sense a parameter is your tweaking constant
anonymous
  • anonymous
:-(
perl
  • perl
the parameter is 'a'*
anonymous
  • anonymous
I thought a was just a slope
Mimi_x3
  • Mimi_x3
Im lost =/ Parameter P(2ap,ap^2)_
perl
  • perl
changing 'a' changes your parabola. but for the line y = mx + b, changing m,b changes your line
perl
  • perl
these are your tweaking values .
perl
  • perl
I dont like their use of p, scratch that and use t instead
perl
  • perl
ok they are using 'parametric equations' , x = f(t), y = g(t)
perl
  • perl
so x = 2at, y = at^2
anonymous
  • anonymous
I like to use matrices and vectors.
Mimi_x3
  • Mimi_x3
Yeah thats in cartesian form
perl
  • perl
ok, use a matrix or vector, id like to see that
perl
  • perl
mimi, when you remove t, you get cartesian equation
Mimi_x3
  • Mimi_x3
matrix or vector ?
Mimi_x3
  • Mimi_x3
Oh yeah my bad i forgot xD
anonymous
  • anonymous
http://img392.imageshack.us/img392/8476/lach11n72fb.jpg
Mimi_x3
  • Mimi_x3
I haven't learnt that complicated stuff yet =/
anonymous
  • anonymous
you will learn it soon.
Mimi_x3
  • Mimi_x3
Hope not looks hard already =/
perl
  • perl
thats curve fitting, no big deal
anonymous
  • anonymous
you will find it easier than conics
Mimi_x3
  • Mimi_x3
Whats conics ?
perl
  • perl
its a quick way to fit a quadratic sequence, cubic, etc
perl
  • perl
mimi, what we are doing is conics, oh dea
anonymous
  • anonymous
conic sections: parabolas, circles, ellipses, hyperbolas, stuff.
perl
  • perl
oh dear *
anonymous
  • anonymous
you do them along parametric equations usually.
perl
  • perl
agdg, put the proper definition on board ;)
anonymous
  • anonymous
well, they're what you get when you cut a cone.
Mimi_x3
  • Mimi_x3
LOL, this is not conics
perl
  • perl
a conic has a specific algebraic equation in x and y , quadratic Ax^2 + Bxy + cY^2 + dy + ex + f = 0
perl
  • perl
yes it is, a parabola is a conic
perl
  • perl
a conic is a family of curves, as agdg pointed out
Mimi_x3
  • Mimi_x3
But Ax^2 + Bxy + cY^2 + dy + ex + f = 0 ?
anonymous
  • anonymous
forget the algebra and xs and ys and f(x)s, you will understand them really well once you picture them geometrically
anonymous
  • anonymous
http://mathworld.wolfram.com/ConicSection.html
perl
  • perl
right
Mimi_x3
  • Mimi_x3
Um, this doesnt look like a conic
perl
  • perl
agdg, whats the nth term of this quadratic sequence 3,5,10,18...
perl
  • perl
lets test your matrix skills
anonymous
  • anonymous
hmmm... is it infinity?
perl
  • perl
no, the nth term
perl
  • perl
give me a formula for the sequence
perl
  • perl
obviously it diverges, ;)
anonymous
  • anonymous
its \[2^n + 2^\frac{n}{2}\]
perl
  • perl
no, lol
anonymous
  • anonymous
wait! it's \[\frac{2^n+2^{\frac{n}{2}}}{2}\]
perl
  • perl
its a quadratic sequence, like in your matrix
anonymous
  • anonymous
how do we find it :-D
anonymous
  • anonymous
so what's the 5th term
perl
  • perl
ok so first we find the differences
perl
  • 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
  • anonymous
that looks like a long matrix
perl
  • perl
3, 5, 10, 18, 29 ... thats the sequence
perl
  • perl
it should fit f(n) = an^2 + bn + c
anonymous
  • anonymous
I guess that works.
anonymous
  • anonymous
how do we multiply matrices?
perl
  • perl
are you asking in general ?

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