a locus question (1 Viewer)

[freaking_out]

I have the following questions:

1.find the equation of the tangent to the curve y^3=27x^2 at the point (t^3,3t^2), hence determine the locus of the point of intersection of perpendicular tangents to the curve.


2. if x^3+3px^2+3qx+r=0 has a double root, show that the double root must be (pq-r)/(2q-2p^2)

thanx in advance.

 

[underthesun]

for the gradient in terms of t:

y = 3t^2;

dy/dt = 6t;

x = t^3;

dx/dt = 3t^2;

dy/dx = (dy/dt)/(dx/dt) = 6t/3t^2 = 2/t;

use line eq.

y - 3t^2 = (2/t) * (x - t^3);
ty - 3t^3 = 2x - 2t^3;
ty = 2x + t^3;

y = 2x/t + t^2;

I got the tangent equation..

now, I don't fully understand the meaning of "perpendicular tangents to the curve".. someone enlighten me?

Q.2 is something to do with multiple roots and differentiation or something.. needs to brush my 4U up sometimes soon

 

[freaking_out]

actually thats how far i got...

but i fink, "perpendicular tangents to the curve" means the perpendicual tangents of the tangent that you just found now
i.e (y = 2x/t + t^2). therefore it is saying, where is this "perpendicular" tangents cross the curve again? if you know what i mean

 

[Harimau]

Originally posted by freaking_out
actually thats how far i got...

but i fink, "perpendicular tangents to the curve" means the perpendicual tangents of the tangent that you just found now
i.e (y = 2x/t + t^2). therefore it is saying, where is this "perpendicular" tangents cross the curve again? if you know what i mean

Which book is it from? Or is it from an exam?

 

[freaking_out]

i think from a txt. book, it was a photocopied sheet i had.

 

[OLDMAN]

I'll help with 1.

As Underthesun derived eqn. of tangent is
y = 2x/t + t^2 (1)
Let u be another parameter point, its tangent eqn is
y = 2x/u + u^2 (2)
For the pt. of intersection, solve for x,
x= (u+t)ut/2
and sub in (1), to get
y=u^2+t^2+ut
But (1) and (2) are perpendicular hence, (2/t)(2/u)=-1
giving ut=-4

y=u^2+t^2+ut=(u+t)^2-ut

Thus a relationship could be obtained purely with x and y,

y=(x^2)/4 +4 a parabola.

 

[underthesun]

I kind-of get it, but does these kind of question ever appear at exams? This is the first time i see a question of such degree of complicacy.

It's a textbook question, so it seems like it's a hard textbook..

 

[freaking_out]

hew, i am glad that this type of crappy questions don't appear in exams.

 

[ND]

Originally posted by underthesun
I kind-of get it, but does these kind of question ever appear at exams? This is the first time i see a question of such degree of complicacy.

It's a textbook question, so it seems like it's a hard textbook..

Isn't this just a 3u question?

 

[ND]

Originally posted by freaking_out


2. if x^3+3px^2+3qx+r=0 has a double root, show that the double root must be (pq-r)/(2q-2p^2)

P(x) = x^3 + 3px^2 + 3qx + r = 0
P'(x) = 3x^2 + 6px + 3q = 0 (because of the double root)
x^2 = -2px - q ...[2]
sub into P(x):
-2px^2 -qx -6p^2x -3pq +3qx + r = 0
2px^2 = -qx -6p^2x -3pq +3qx + r ...[3]
now [3] = [2]*2p:
-4p^2x -2pq = -qx - 6p^2x -3pq +3qx + r
2p^2x - 2qx = r - pq
x(2q-2p^2) = pq-r
x = (pq-r)/(2q-2p^2)

 

[underthesun]

Originally posted by ND


Isn't this just a 3u question?

yah, parabola parametrics, but this perpendicular tangent intersection, never seen in 3u textbooks.

thank god polynomials isn't on my upcoming test

 

[OLDMAN]

ND :Isn't this just a 3u question?

freaking_out: hew, i am glad that this type of crappy questions don't appear in exams.

Pity the poor question!

Yes it does have a retro feel to it, like Ye Olde 4 Unit, circa 1979. However even if it is just the lowly parabola- heavens whatsit doing here?!- there's a few valuable learning points to it.

eg.

1) Locus questions are almost invariably either point of intersection of two lines or midpoint of two points. So you need two lines to get an intersection
2)A derived locus equation do not contain any moving parameter eg. u,t or p,q.
3) Almost invariably, getting rid of these involve algebraic manipulation of (t+u) and (uv) without eliminating one variable in terms of the other, though last year's rec. hyperbola locus question actually involved the elimination of p in terms of q.

 

[freaking_out]

Isn't this just a 3u question?

if it is a 3u question, then GOD SAVE US ALL!!!

 

[underthesun]

if it is a 3u question, then GOD SAVE US ALL!!!

harder 3u question includes 3u questions.

This question is harder than 3u questions.

But it is a 3u questions.

Then it is a harder 3u question.

Then it is included in 4unit

 

[ND]

Originally posted by underthesun


harder 3u question includes 3u questions.

This question is harder than 3u questions.

But it is a 3u questions.

Then it is a harder 3u question.

Then it is included in 4unit

The 3u textbook that we use has quite a few of these perpendicular tangent type q's, which leads me to believe that this q. was just a standard 3u question. I'm not really sure on the difficulty of the 3u paper, but was this question even hard enough for q7?

 

[underthesun]

Because your textbook has lots of perp.tangent intersection questions on it, it make it easier for it's users. But for people who even hardly heard of perp. tangents (like me), it seems really hard. Experience is in need in these kinds of questions i guess..

what textbook are you using?
(I use excel and T.Lee)

 

[ND]

For 4u i use mainly the T.Lee, but i also have the Fitzpatric and Excel for revision.