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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.
 2 years ago
 2 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.
 2 years ago
 2 years ago

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perlBest ResponseYou've already chosen the best response.1
mmmmm, you werent kidding around
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Is this easy for you ? xD
 2 years ago

perlBest ResponseYou've already chosen the best response.1
i will have to think about it
 2 years ago

perlBest ResponseYou've already chosen the best response.1
the directrix is that line under the parabola
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
lols, you were thinking that its easy ? xD
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
I live in Australia, so its kinda different, there's only a maths class xD
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Yes I have, how did you know ? xD
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
What grade are you in ?
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Cool~ college, and you havent learnt that ?
 2 years ago

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

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

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Yeah, I don't really know where's the external point is, thats why im lost =/
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Something special ? What do you mean ?
 2 years ago

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

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

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

perlBest ResponseYou've already chosen the best response.1
where the tangents meet the parabola, the two tangents
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
wow~ nice pdf file, its detailed on parametrics ty
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
But where is the external point ?
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Can't it have to be specific
 2 years ago

perlBest ResponseYou've already chosen the best response.1
in the pdf, the point is below the parabola
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
It can be but it doesn't make sense when it says RT subtends a right angle at the focus.
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
I know but where's the external point ?
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Is the parabola imaginery or somethinbg ?
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
I don't get your diagrams ? sorry
 2 years ago

perlBest ResponseYou've already chosen the best response.1
yeah the diagram is a beast
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
lols, why is there like 3 digrams ? i dont get it >_<
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
lols, this is not 3D trig xD
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Nope, i dont know how to start, the problem is where is T
 2 years ago

perlBest ResponseYou've already chosen the best response.1
just a point below the parabola
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
But it won't make sense, and i dont know how to prove
 2 years ago

perlBest ResponseYou've already chosen the best response.1
i think they mean the segment R T
 2 years ago

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

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

Mimi_x3Best ResponseYou've already chosen the best response.4
idk, never mind then i will work it out myself
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Yeah i know, none of the hard questions have diagrams =/
 2 years ago

perlBest ResponseYou've already chosen the best response.1
i think RT is the line segment though
 2 years ago

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

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Yeah, but im having problem proving
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
lols, you got it from boredofstudies right ?
 2 years ago

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

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Yeah but the problem is that I don't understand the solution
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
it helps to draw a diagram.
 2 years ago

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

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

Mimi_x3Best ResponseYou've already chosen the best response.4
But what's xx1 ? im trying to figure it out
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
oh i get it, but where did y=1/2(p+q)xapq come from ?
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
where did 2a come from ?
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
no, isn't it x^2 = 4ay y = x^2/4a y' = 2x/4a = x/2a
 2 years ago

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

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
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
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
oh the equation of the tangent to the parabola is y=pxap^2 right ?
 2 years ago

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

perlBest ResponseYou've already chosen the best response.1
the chords of tangents formula
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
now it all makes sense
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Doesn't it involve parameters as well P(2ap,ap^2)
 2 years ago

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

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

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Im confused, is parameters involved ?
 2 years ago

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

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

perlBest 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
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
isn't parameters P(2ap,ap^2) So there must be a point for the parameters ?
 2 years ago

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

perlBest ResponseYou've already chosen the best response.1
right, but they are considered 'parameters'
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
but its not a straight line so it cant be y=mx + b
 2 years ago

perlBest ResponseYou've already chosen the best response.1
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
 2 years ago

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

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
I thought a was just a slope
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Im lost =/ Parameter P(2ap,ap^2)_
 2 years ago

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

perlBest ResponseYou've already chosen the best response.1
these are your tweaking values .
 2 years ago

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

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

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
I like to use matrices and vectors.
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Yeah thats in cartesian form
 2 years ago

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

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

Mimi_x3Best ResponseYou've already chosen the best response.4
Oh yeah my bad i forgot xD
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
http://img392.imageshack.us/img392/8476/lach11n72fb.jpg
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
I haven't learnt that complicated stuff yet =/
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
you will learn it soon.
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Hope not looks hard already =/
 2 years ago

perlBest ResponseYou've already chosen the best response.1
thats curve fitting, no big deal
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
you will find it easier than conics
 2 years ago

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

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

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

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
you do them along parametric equations usually.
 2 years ago

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

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

Mimi_x3Best ResponseYou've already chosen the best response.4
LOL, this is not conics
 2 years ago

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

perlBest ResponseYou've already chosen the best response.1
yes it is, a parabola is a conic
 2 years ago

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

Mimi_x3Best ResponseYou've already chosen the best response.4
But Ax^2 + Bxy + cY^2 + dy + ex + f = 0 ?
 2 years ago

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

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
http://mathworld.wolfram.com/ConicSection.html
 2 years ago

Mimi_x3Best ResponseYou've already chosen the best response.4
Um, this doesnt look like a conic
 2 years ago

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

perlBest ResponseYou've already chosen the best response.1
lets test your matrix skills
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
hmmm... is it infinity?
 2 years ago

perlBest ResponseYou've already chosen the best response.1
give me a formula for the sequence
 2 years ago

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

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

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

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
how do we find it :D
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
so what's the 5th term
 2 years ago

perlBest ResponseYou've already chosen the best response.1
ok so first we find the differences
 2 years ago

perlBest ResponseYou've already chosen the best response.1
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
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
that looks like a long matrix
 2 years ago

perlBest ResponseYou've already chosen the best response.1
3, 5, 10, 18, 29 ... thats the sequence
 2 years ago

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

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
I guess that works.
 2 years ago

agdgdgdgwngoBest ResponseYou've already chosen the best response.0
how do we multiply matrices?
 2 years ago

perlBest ResponseYou've already chosen the best response.1
are you asking in general ?
 2 years ago
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