Parametrics and the parabola marathon. (1 Viewer)

[anomalousdecay]

I have posted this thread in light that everyone will benefit from practising these HSC questions.
I tend to have problems with the locus questions of this topic.

To start off:

2005 HSC question 4(c). (5 marks altogether)

The points P(2ap, ap^2) and Q(2aq, aq^2) lie on the parabola x^2 = 4ay.
[The equation of the normal to the parabola at P is x +py = 2ap + ap^3.
The equation of the normal at Q is similarly given by x + qy = 2aq + aq^3.


(-apq(p+q), a( (p^2) + pg + (q^2)+2 )






I could only get the second part.

 

[HeroicPandas]

In locus they always tell u to find a point, in this case it is

R(-apq(p+q), a^2 (p^2 + pq + q^2 +2))

From part (ii), we got pq = -1 (because PQ is focal chord)

Now lets write out R

x = -apq(p+q)

x = a(p+q) (as pq = -1)

y = a^2(p^2 + pq + q^2 +2)

y = a^2(p^2 + q^2 + 1)

We need to get rid of parameters: p and q
We realy need a link between x and y, so i notice (p^2 + q^2) having the ability to be transformed into (p+q)^2 somehow....and so i can make (p+q) from the x-coordinate the subject and substitute it into the transformed y-coordinate

So we do: (p+q)^2 - 2pq = p^2 + q^2

Therefore, y = a^2 [ (p+q)^2 - 2pq]

y = a^2 [ (p+q)^2 + 2] (as pq = -1)


Now,

x = a(p+q)
y = a^2 ((p+q)^2 + 2)

and u will Know what to do

 

[anomalousdecay]

So then is it y = x^2 + 2a^2?

 

[HeroicPandas]

So then is it y = x^2 + 2a^2?

yeh sometimes depending on my fear i would say "therefore the locus of R is the parabola ______ with focus (__,__)"

 

[anomalousdecay]

Yeah that's true. I just do Vertex as it is just simply (0, 2a^2). If you do Focus, It is an extra step. May as well add that eccentricity equals 1

 

[anomalousdecay]

Next question. 2001 HSC 6(b).

 

[HeroicPandas]

Next question. 2001 HSC 6(b).

Find a connection between the point x and y by means of squaring, using perfect square factorisation, etc

(its so hard to explain the way of thinking for this question without giving it away)

 

[HeroicPandas]

Yeah that's true. I just do Vertex as it is just simply (0, 2a^2). If you do Focus, It is an extra step. May as well add that eccentricity equals 1

thats overkill lol

 

[anomalousdecay]

Find a connection between the point x and y by means of squaring, using perfect square factorisation, etc

(its so hard to explain the way of thinking for this question without giving it away)

I got stuck at (ii) and you can't really try the rest of the question without (ii).

 

[HeroicPandas]

I got stuck at (ii) and you can't really try the rest of the question without (ii).

m(gradient P) = 1/t

.: m(gradient Q) = -t (as normal Q is perpendicular to normal P)





We know that the gradient of normal Q is -t





Sub this back into parabola as it lies on it and u get: y = at^2

SO,

Q[-2at, at^2]

 

[anomalousdecay]

I did 2009 2(c).
(i) is straight foward and easy.
I got R(3t, 2t^2) for (ii) and the tangent at q: y = 2tx - 4t^2
For (iii) I got the locus is y = 2x^2/9 .

 

[anomalousdecay]

m(gradient P) = 1/t

.: m(gradient Q) = -t (as normal Q is perpendicular to normal P)





We know that the gradient of normal Q is -t





Sub this back into parabola as it lies on it and u get: y = at^2

SO,

Q[-2at, at^2]

isn't the gradient at p -1/t?

 

[HeroicPandas]

I did 2009 2(c).
(i) is straight foward and easy.
I got R(3t, 2t^2) for (ii) and the tangent at q: y = 2tx - 4t^2
For (iii) I got the locus is y = 2x^2/9 .

Nice

 

[HeroicPandas]

isn't the gradient at p -1/t?

u are right, sorry

 

[anomalousdecay]

For 6(b) I skipped (iii).

For (iv) I got the locus is a parabola of equation x^2 = a(y-3a) with vertex (0,3a) and Focus (0, 13a/4).

 

[HeroicPandas]

For 6(b) I skipped (iii).

For (iv) I got the locus is a parabola of equation x^2 = a(y-3a) with vertex (0,3a) and Focus (0, 13a/4).

yeh got that too

 

[HeroicPandas]

there was some confusion with the gradient thing lol

m(norm. P) = -1/t
.: m(norm. Q) = t
.: m(tang. Q) = -1/t

dy/dx = x/2a

For the point of contact of tangents with gradient -1/t

-1/t = x/2a
x = -2a/t
y = 2a/t^2

Q[-2a/t, 2a/t^2]

(in my previous try, i subbed in dy/dx as gradient of norm. Q which was wrong because dy/dx represents gradient not normal)

 

[anomalousdecay]

I'll try and ask my teacher that question. If anything will cost me a band 6, it will be a shock 9 marker like that.
Got to practice this stuff a lot.