Conics qu (2 Viewers)

[Giant Lobster]

damnit i should be able to do these questions, after all they are from my crappy school im so pissed at myself and the world in general grrrr >.<

anyway

H is the hyperbola 2xy = a^2 with P(at/2 , a/t) as parameter
from part a) u get the tangent eq as 2x + t^2 = 2at

S(a,a) is a point. ST is the perpendicular to the tangent at P, T is on the tangent. Prove locus of T is a circle, as P varies.

I did it the normal way (no tricks) and couldnt resolve t. Algebra too gay. I dont get it, what other method can be used?
thanks

 

[ND]

I don't think you need any tricks. You know the eqn of ST (cos you know a pt it passes through and it's gradient), and you know the eqn of the tangent; solve simultaneously.

edit: btw, you forgot the z position in your location.

 

[McLake]

If ST is perp to tangent at P, and T is on tangent and ST, then T IS P. Are you sure that that is correctly worded?

 

[Giant Lobster]

???
Yeah its correctly worded
I get the diagram, i just need someone to verify that the algebra is impossible (or thereabouts)

P is some moveable point
S is a constant point (a,a)
A perpendicular to the tangent at P is drawn from S. T is the foot of this perpendicular. T does not necessarily equal P. Sorry if I initially worded it wrong

can someone help, pls. Im really worried I fully underestimated my school's math assessment making skills

 

[ND]

Originally posted by McLake
If ST is perp to tangent at P, and T is on tangent and ST, then T IS P. Are you sure that that is correctly worded?

Not necessarily. See the attached image.

 

[Giant Lobster]

*bump

i still have yet to be helped >.<
thx

 

[McLake]

My bad, I read wrong.

Answer:

2xy = a^2
y = a^2/(2x)
dy/dx = -a^2/2(x^2)
at at/2
m = -a^2/2(a^2t^2/4)
m = -1/(t^2/2)
m = -2/t^2

so y - a/t = -2/t^2(x - at/2)
y = -2x/t^2 - 2a/t
so
2x + t^2*y = 2at

now m(norm) is t^2/2

so y - a = t^2/2(x - a)
y = t^2*x/2 - a*t^2/2 + a
so ST: y = t^2*x/2 - (a*t^2 - 2a)/2

solve sim
2x + t^2(t^2*x/2 - (a*t^2 - 2a)/2) = 2at
2x + t^4x/2 - (a*t^4 - 2a*t^2)/2 = 2at
4x + t^4x - a*t^4 + 2a*t^2 = 4at
x(4 - t^4) = 2at(2 - t)
x = 2at(2 - t)/(2 - t^2)(2 + t^2)

DAMN

 

[Giant Lobster]

see? gay tedious work
wot kind of 4u question is that
doesnt test skills or anything, just how careful u are and patience (unless one can find a better way)
thanx for trying

 

[hyparzero]

Ok, sorry to bring back an old post, but the question hasn't seemed to have been solved yet. I got up to the step where McLake simultaneously solved ST with the tangent at P.

However, i got the Locus of T as:
4x³ - 4ax² + 2y³ - 2ay² - 8ay + 8xy² -8axy + 8a² = 0

Which resembles some sort of type of general elliptical function... need some maths genius to fix the question... anyone care to help?

 

[Riviet]

I had a go at the question and got up to the point where I found the co-ordinates of T in terms of a and t. Then I was stuck, as the numerator of each co-ordinate had all kinds of t terms (t, t², t³). I was thinking there would be some restriction that we could find involving t?

hyparzero: I'd like to know how you managed to eliminate t and get that locus.

 

[hyparzero]

I had a go at the question and got up to the point where I found the co-ordinates of T in terms of a and t. Then I was stuck, as the numerator of each co-ordinate had all kinds of t terms (t, t², t³). I was thinking there would be some restriction that we could find involving t?

hyparzero: I'd like to know how you managed to eliminate t and get that locus.

I let t = 1 and viola! t simply disappears!

kidding,
look at the following line:
y = t²x/2 - (at² - 2a)/2 .....Equation of ST

Simply rewrite it by taking t² as the common factor.
So you'll have
t² = ............and so on

thus t = Root(...........and so on)
plug that back into the tangent equation, then square both sides to get rid of the roots.

 

[sasquatch]

You cant take the square root of t as there can then be both positive and negative solutions. What you can do though is square the tangent equation:

(the original dude wrote it wrong)

2x + t²y = 2at

4x² + 4xat² + t⁴y² = 4a²t²

Now you can subsitute t² in

 

[hyparzero]

You cant take the square root of t as there can then be both positive and negative solutions. What you can do though is square the tangent equation:

(the original dude wrote it wrong)

2x + t²y = 2at

4x² + 4xat² + t⁴y² = 4a²t²

Now you can subsitute t² in

you are amazing, thanks for that