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Mimi_x3
 4 years ago
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
 4 years ago
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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perl
 4 years ago
Best ResponseYou've already chosen the best response.1mmmmm, you werent kidding around

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Is this easy for you ? xD

perl
 4 years ago
Best ResponseYou've already chosen the best response.1i will have to think about it

perl
 4 years ago
Best ResponseYou've already chosen the best response.1the directrix is that line under the parabola

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4lols, you were thinking that its easy ? xD

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4I live in Australia, so its kinda different, there's only a maths class xD

perl
 4 years ago
Best ResponseYou've already chosen the best response.1sorry wrong question, have you done calculus

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Yes I have, how did you know ? xD

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4What grade are you in ?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Cool~ college, and you havent learnt that ?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1not specifically, but i bet i can do this

perl
 4 years ago
Best ResponseYou've already chosen the best response.1im trying to make a diagram, im not even sure i understand the question yet

perl
 4 years ago
Best ResponseYou've already chosen the best response.1ok you have an external pt, and you want the tangent through it

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Yeah, I don't really know where's the external point is, thats why im lost =/

perl
 4 years ago
Best ResponseYou've already chosen the best response.1ok first thing is first. the chord of contact means something special

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Something special ? What do you mean ?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1there is a picture of a chord of contacts http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

perl
 4 years ago
Best ResponseYou've already chosen the best response.1well given an external point there are two tangents from that point to the parabola

perl
 4 years ago
Best ResponseYou've already chosen the best response.1now where the tangents intersect the parabola, now draw a chord through those two points

perl
 4 years ago
Best ResponseYou've already chosen the best response.1where the tangents meet the parabola, the two tangents

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4wow~ nice pdf file, its detailed on parametrics ty

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4But where is the external point ?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Can't it have to be specific

perl
 4 years ago
Best ResponseYou've already chosen the best response.1in the pdf, the point is below the parabola

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4It can be but it doesn't make sense when it says RT subtends a right angle at the focus.

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4I know but where's the external point ?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Is the parabola imaginery or somethinbg ?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4I don't get your diagrams ? sorry

perl
 4 years ago
Best ResponseYou've already chosen the best response.1yeah the diagram is a beast

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4lols, why is there like 3 digrams ? i dont get it >_<

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4lols, this is not 3D trig xD

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Nope, i dont know how to start, the problem is where is T

perl
 4 years ago
Best ResponseYou've already chosen the best response.1just a point below the parabola

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4But it won't make sense, and i dont know how to prove

perl
 4 years ago
Best ResponseYou've already chosen the best response.1i think they mean the segment R T

perl
 4 years ago
Best ResponseYou've already chosen the best response.1yes have faith when you draw it that it will all come into place, the problem will unravel

perl
 4 years ago
Best ResponseYou've already chosen the best response.1i am atheist, what did you think i mean ?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4idk, never mind then i will work it out myself

perl
 4 years ago
Best ResponseYou've already chosen the best response.1would be nice if they gave you a diagram

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Yeah i know, none of the hard questions have diagrams =/

perl
 4 years ago
Best ResponseYou've already chosen the best response.1i think RT is the line segment though

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0\[\text{An equation of the parabola with focus }(0,p)\text{ and directrix }y = p\text{ is}\] \[x^2=4py\]

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0so the directrix would be y = a in your question :D

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Yeah, but im having problem proving

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0If 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/(ay1) d/dx TS=(y1a)/x1 d/dx RS * d/dx TS= 1

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4lols, you got it from boredofstudies right ?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1how did you get chord of tangents equation

perl
 4 years ago
Best ResponseYou've already chosen the best response.1hey, can you draw a diagram then of whats going on, please post the link

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Yeah but the problem is that I don't understand the solution

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0it helps to draw a diagram.

perl
 4 years ago
Best ResponseYou've already chosen the best response.1can you post the link to the solution please , so i can read it

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0http://community.boredofstudies.org/showthread.php?t=192825&page=1

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4But what's xx1 ? im trying to figure it out

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0http://wiki.answers.com/Q/Find_the_equation_of_chord_of_contact_tangent_to_the_parabola

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4oh i get it, but where did y=1/2(p+q)xapq come from ?

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0xx1 must mean something like \[x_1\times x= 2a(y+y_1)\]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4where did 2a come from ?

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0\[parabola: x^2 = 4py\]\[a = \frac{1}{4p}\]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4no, isn't it x^2 = 4ay y = x^2/4a y' = 2x/4a = x/2a

perl
 4 years ago
Best ResponseYou've already chosen the best response.1i wish there were other colors, but from T draw two tangents to the parabola

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0perl'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
 4 years ago
Best ResponseYou've already chosen the best response.4oh the equation of the tangent to the parabola is y=pxap^2 right ?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1they are given here http://www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

perl
 4 years ago
Best ResponseYou've already chosen the best response.1the chords of tangents formula

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0now it all makes sense

perl
 4 years ago
Best ResponseYou've already chosen the best response.1i dont get that parametrisation, is it because they want x^2 = 4ay ? , what is p

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Doesn't it involve parameters as well P(2ap,ap^2)

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0they want something like the quotient of the slopes between 2 lines is equal to 1

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0and one of those lines should be RT

perl
 4 years ago
Best ResponseYou've already chosen the best response.1crap, it missed the R it is the right bottom corner

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Im confused, is parameters involved ?

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0no parameter seems to be involved :(

perl
 4 years ago
Best ResponseYou've already chosen the best response.1actually a is the parameter, you have to be careful, this word parameter is used in different senses

perl
 4 years ago
Best ResponseYou've already chosen the best response.1, 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
 4 years ago
Best ResponseYou've already chosen the best response.4isn't parameters P(2ap,ap^2) So there must be a point for the parameters ?

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0meh, I assumed m and b to be constant

perl
 4 years ago
Best ResponseYou've already chosen the best response.1right, but they are considered 'parameters'

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4but its not a straight line so it cant be y=mx + b

perl
 4 years ago
Best ResponseYou've already chosen the best response.1in 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
 4 years ago
Best ResponseYou've already chosen the best response.1think of in this sense a parameter is your tweaking constant

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0I thought a was just a slope

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Im lost =/ Parameter P(2ap,ap^2)_

perl
 4 years ago
Best ResponseYou've already chosen the best response.1changing 'a' changes your parabola. but for the line y = mx + b, changing m,b changes your line

perl
 4 years ago
Best ResponseYou've already chosen the best response.1these are your tweaking values .

perl
 4 years ago
Best ResponseYou've already chosen the best response.1I dont like their use of p, scratch that and use t instead

perl
 4 years ago
Best ResponseYou've already chosen the best response.1ok they are using 'parametric equations' , x = f(t), y = g(t)

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0I like to use matrices and vectors.

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Yeah thats in cartesian form

perl
 4 years ago
Best ResponseYou've already chosen the best response.1ok, use a matrix or vector, id like to see that

perl
 4 years ago
Best ResponseYou've already chosen the best response.1mimi, when you remove t, you get cartesian equation

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Oh yeah my bad i forgot xD

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4I haven't learnt that complicated stuff yet =/

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0you will learn it soon.

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Hope not looks hard already =/

perl
 4 years ago
Best ResponseYou've already chosen the best response.1thats curve fitting, no big deal

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0you will find it easier than conics

perl
 4 years ago
Best ResponseYou've already chosen the best response.1its a quick way to fit a quadratic sequence, cubic, etc

perl
 4 years ago
Best ResponseYou've already chosen the best response.1mimi, what we are doing is conics, oh dea

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0conic sections: parabolas, circles, ellipses, hyperbolas, stuff.

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0you do them along parametric equations usually.

perl
 4 years ago
Best ResponseYou've already chosen the best response.1agdg, put the proper definition on board ;)

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0well, they're what you get when you cut a cone.

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4LOL, this is not conics

perl
 4 years ago
Best ResponseYou've already chosen the best response.1a conic has a specific algebraic equation in x and y , quadratic Ax^2 + Bxy + cY^2 + dy + ex + f = 0

perl
 4 years ago
Best ResponseYou've already chosen the best response.1yes it is, a parabola is a conic

perl
 4 years ago
Best ResponseYou've already chosen the best response.1a conic is a family of curves, as agdg pointed out

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4But Ax^2 + Bxy + cY^2 + dy + ex + f = 0 ?

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0forget the algebra and xs and ys and f(x)s, you will understand them really well once you picture them geometrically

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.4Um, this doesnt look like a conic

perl
 4 years ago
Best ResponseYou've already chosen the best response.1agdg, whats the nth term of this quadratic sequence 3,5,10,18...

perl
 4 years ago
Best ResponseYou've already chosen the best response.1lets test your matrix skills

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0hmmm... is it infinity?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1give me a formula for the sequence

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0its \[2^n + 2^\frac{n}{2}\]

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0wait! it's \[\frac{2^n+2^{\frac{n}{2}}}{2}\]

perl
 4 years ago
Best ResponseYou've already chosen the best response.1its a quadratic sequence, like in your matrix

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0how do we find it :D

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0so what's the 5th term

perl
 4 years ago
Best ResponseYou've already chosen the best response.1ok so first we find the differences

perl
 4 years ago
Best ResponseYou've already chosen the best response.1the 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

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0that looks like a long matrix

perl
 4 years ago
Best ResponseYou've already chosen the best response.13, 5, 10, 18, 29 ... thats the sequence

perl
 4 years ago
Best ResponseYou've already chosen the best response.1it should fit f(n) = an^2 + bn + c

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0I guess that works.

agdgdgdgwngo
 4 years ago
Best ResponseYou've already chosen the best response.0how do we multiply matrices?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1are you asking in general ?
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