Parametrics Questions (1 Viewer)

[CM_Tutor]

In another thread, nike33 asked for some more interesting parametrics questions, so here we go...

1. P(2ap, ap²) and Q(2aq, aq²) are two points on the parabola x² = 4ay. The tangent at P and the line through Q parallel to the axis of the parabola meet at R, and the tangent at Q and the line through P parallel to the axis of the parabola meet at S. Show that PQRS is a parallelogram, and show that its area is 2a²|p - q|³.

2. The points P(2ap, ap²) and Q(2aq, aq²) both lie on the parabola x² = 4ay. The tangents at P and Q meet at T, such that angle PTQ is 60. Find the locus of T.

3. --- deleted --- This question was wrong. The correct one is posted further down this thread.

4. R(2ar, ar²) lies on the parabola x² = 4ay. Perpendiculars are drawn from R to the x- and y- axes, meeting them (respectively) at X and Y. If M is the midpoint of RY and T is the midpoint of MX, show that the locus of T is a parabola, and find its equation.

5. The point P(2a, a²), a > 0, lies on the parabola y = x² / 4. The tangent at P meet the x-axis at X, and S is the focus of the parabola. If angle SPX is alpha, and the acute angle between PX and the x-axis is beta. By finding values for tan alpha and tan beta, or otherwise, find the value of alpha + beta, and the coordinates of P if alpha = beta.

6. The point P lies on the parabola x² = 4ay, which has a focus at S. A line is drawn through the vertex of the parabola (O), parallel to the tangent at P, and it meets the parabola again at Q. The tangents at P and Q meet at R. Show that the locus of R is a parabola, and if this parabola has focus at S', show that S divides the interval OS' in the ratio 8:1.

I have more, but that's probably enough for now...

 

[nike33]

1.

m_pq = (p+q) / 2

Finding R

y - px + ap² = 0 (1)
x = 2aq (2)

sub (2) into (1)

y - 2apq + ap² = 0
y = 2apq - ap²

Hence R (2aq, 2apq - ap²)
S(2ap,2apq - aq²) {Similarly)

m_rs = [2apq - aq² - 2apq + ap²] / (2aq - 2ap)
= (p-q)(p+q) / 2(p-q)
= (p+q)/2

PQ||RS
QR||PS

.:. angleQRP = angleRPS
.:. angleSRP = angleRPQ

RP is common

hence by AAS triangle QPR is congruent to triangle PRS
therefore PQRS is a parallelogram (Diagonals of a parallelogram bisect it into two congruent triangles)


[source CM_Tutor]

The distance between PS and QR is 2a|p - q|, and so A_PQRS =
2a|p - q| * a(p - q)² = 2a²|p - q|³, as required.

 

[nike33]

2.

T[a(p+q), apq]

x = a(p+q)
y = apq

Let Z and Y be the points where the tangents and Q and P respectively cut the x axis

angleQZX = tan^-1q
anglePYX = tan^-1p
angleTYZ = tan^-1p (vert. opp)

.:. pi/3 + tan^-1p = tan^-1q (sum of opposite interior angles of triangle = exterior angle)

tan pi/3 = tan(tan^-1q - tan^-1p)
sqr(3) = (q-p)(1 + pq)
p-q = sqr(3) + pq.sqr(3)

now x = a(p+q)
x²= a²(p+q)²
= a²(p² - 2pq+q² + 4pq)
= a²((q-p)² + 4y / a) as y/a = pq
= a²((sqr(3) + sqr(3).pq)² + 4y / a)
= a²( 3 + 6pq + 3(pq)² + 4y / a)
= a²( 3 + 6y/a + 3(y/a)² + 4y/a)
= 3a² + 10ay + 3y²

 

[nike33]

5. "Otherwise"

let
B = beta
A = alpha

let m be where the tangent at p cuts the y axis

angle(smp) = A (The tangent at a pt P on a parabola is equally inclined to the axis of the parabola and the focal chord through P

angle(MXx) = B (vert opp..) (where x is at - infinity on x axis)

A + B + 90 = 180 (sum of triangle)
A + B = pi/2

let S(o,p) be the focus

at a² = p
a = sqr(P) as a>0

hence P(2sqr(P), p)

"Using the tan method"

Obviously tanB = a
using the angle between two lines method we get tanA = 1/a

hence
TanBTanA = 1
TanB = Tan(pi/2 - A)
B + A = pi/2

 

[CM_Tutor]

Originally posted by nike33
1.

...edited out...

Area = PS.(P to the line QR where distance is minimised) (absolute value as negative values may occur)
= |(ap² - 2apq - aq²)(2ap + 2aq)|
= |-2a²(q+p)³|
= 2a²|(q-p)³|

I don't follow the last steps here. How did |(q + p)³| suddenly become |q - p|³.

I'll look at the rest tonight - the source is not with me at the moment, and I have a lot of other things to deal with today.

 

[Xayma]

Hmm for question 3 I think A and B are the points of intersection between the lines and the directix.

 

[CM_Tutor]

Question 1: Nike33, there is an easier way. If you conside how PS and QR are constructed, both are parallel to the axis, and hence are necessarily parallel to one another and vertical.

Thus, it is easily shown that PS = QR = a(p - q)². It follows that PQRS is a parallelogram (one pair of opposite sides equal and parallel).
The distance between PS and QR is 2a|p - q|, and so A_PQRS = 2a|p - q| * a(p - q)² = 2a²|p - q|³, as required.

Question 2 is correct so far, but how would you describe the locus?

Question 3. Well, where do I start. I have no idea what I was thinking when I typed it, as it's not even close to the question I intened, as the whole end bit (the interesting bit) is missing! I'll edit the above post to take it out, and here is the actual question:

3. P(2ap, ap²) and Q(2aq, aq²) are two points on the parabola x² = 4ay such that PQ is a focal chord. S is the focus of the parabola, and A and B are (respectively) the feet of the perpendiculars from P and Q to the directrix. If M is the midpoint of AB, show that MS and PQ are perpendicular. Hence, or otherwise, show that the ratio of the area of triangle PQM to the area enclosed by the parabola and the line PQ is independent of p and q.

We don't have an answer to 4 yet, nor to 6, although nike33 is on the right track, so I'll leave them for the moment.

Question 5 - I have a couple of questions:

1. How do you know that angle SMP is A? I know it's true, but how do you know it?

2. What is p?

I agree that this is fairly straight-forward - provided you 'otherwise' it. The indicated method is not as nice.

PS - Have a look at the 1993 2u HSC, question 10(b). Can you find a proof that the result it proves is general? ie, if A and B lie on a parabola, such that the vertex V lies on the parabola between A and B, and P is any point on the parabola between A and B, then the maximum possible area of triangle APB is three-quarters of the area of parabolic segment APB?

 

[nike33]

ahh ill edit and fix them up for anyone thats interested in the answers

 

[CM_Tutor]

Nike33, I had a typo in my formula for the area in question 1 - I had a cubed where I meant a squared, which I have corrected. Could you correct it where you have quoted me in your question 1 answer? Thanks.

 

[nike33]

4. M(ar, ar²)
X(2ar,0)

.:. T ((3ar)/2, (ar²)/2)

x = 3ar / 2
y = ar² / 2

elimintate r and x² = (9/2)ay

 

[nike33]

6.

P(2ap,ap²) Q(2aq,aq²) S(0,a)

M_oq = (aq²) / 2aq
= q / 2
= p (as OQ is parallel to tangent at p)

q = 2p

hence Q(4ap,4ap²)
tangent of Q has gradient 2p
and equation y - 2px + 4ap² = 0

now y - 2px + 4ap² = 0 ___1
and y - px + ap² = 0 ___2

1 - 2 eliminating y

and you get x = 3ap, sub this back in and get

R(3ap,2ap²)

solve these parameters and x² = (9/2) ay

ie a parabola with Focus S' (0, (9/8)a)


//NB dont do this in an exam!

sub in m=8 n =1

S(x,y)
x = 0 (obviously)
y = [(m.(9/8)a + n0) / (m + n)]
= 9a / 9
= a

hence S(0,a) which is true, .:. S divided OS' in ratio 8:1

 

[nike33]

3. M(a(p+q),-a) {find B, A and then M = (B + A) / 2}

Gradient SM -> -a-a / a(p+q) = -2/ (p+q)
Gradient PQ -> p+q / 2

m_sm * m_pq = -1

hence MS _|_ QP

"Otherwise"

notice that M represents the intersection of tangents from P and Q. ie a(p+q),apq but pq = -1 hence a(p+q), -a ...which is M

obviously the ratio of areas is independant of p and q, as the area of the parabola enclosed by the line PQ just exactly fits in the triangle QPM and will remain this way in a fixed ratio for all values of the parameters p and q

 

[nike33]

hey CM is there any elegant method of doing this qn (out of cambridge) because it took me like 25 mins and 4.5 pages

P(2ap,ap²), X(2ax, ax² ) Z(2az, az²) are variable points on x² = 4ay. Suppose that T is the point of intersection of the tangents to the parabola at X and Z

a) Show T has co-ordinates (a(x + z), axz)
b)Show that the line through T perpindicular to the tangent at P meets the directx at the point D (a(p+x+z+pxz), -a)

now i dont think there would be a fast method of those two..but this c) took me ages and im wondering if theres a geometrical approach or something maybe

c) Hence (ill add or otherwise ) or otherwise find the locus of the orthocentre of the triangle fromed by the three tangents to the parabola drawn at P X Z

NB soln of locus is y = -a.... after all that algebra bashing it came down to this

if you have any other 'unusual' locus/ parabola qns please post! unusual as in no-one in their right mind would have seen them before..hehe

 

[CM_Tutor]

Originally posted by nike33
"Otherwise"

notice that M represents the intersection of tangents from P and Q. ie a(p+q),apq but pq = -1 hence a(p+q), -a ...which is M

obviously the ratio of areas is independant of p and q, as the area of the parabola enclosed by the line PQ just exactly fits in the triangle QPM and will remain this way in a fixed ratio for all values of the parameters p and q

I agree with the first bit, but I don't follow the second. Why does "the area of the parabola enclosed by the line PQ just exactly fits in the triangle QPM"? - In fact, I didn't get the ratio as 1:1, but I could have made an algebra error. Could you expand on this answer?

BTW, I'll have a look at your Cambridge question tonight, and will get back to you. Also, are you doing / have you done the extra projectiles questions I posted?

 

[nike33]

for the second part...im so bad at explaining things, but im not saying the ratio is 1:1 but it must be a fixed ratio independant of p and q as P and Q are similar in both the area of the parabola and the triangle and when in a ratio cancel out..

.arrg this is confusing me but it just is..ie independant of p and q drawing the triangle PQM where P and M are tangents to the parabola by observation ahh i hate this

(and yes i did the projectile motion qns ill post my answers if no-one else has a go before the next couple of days

 

[CM_Tutor]

Nike33, I've been thinking about this - occasionally, I have a huge fortnight at the moment - and I still don't see the geometric argument that you are using. Perhaps you could explain some more? - for example, suppose this were an exam question. What would you write. I agree that M must be the intersection of the tangents from P and Q, but I don't follow where you went from there. Are you saying that all possible PQM's are similar? If so, why?

My answer shows the ratio dependent on a, but not p or q - is this consistent with what you are thinking?

PS: Sorry I didn't get to the Cambridge question - I'll have a look tonight. Promise.

 

[nike33]

haha i cant explain..but its ..just has to

My answer shows the ratio dependent on a, but not p or q - is this consistent with what you are thinking?

yes this is what im saying, and im saying its true because the line PQ is similar for both the triangle and the area of parabola.. draw any triangle PQM, and the parabola under PQ just exactly fits inside it hence the only factor affecting their area ratios is the focal length 'a' and not p/ q..

in an exam i would have just used the 'hence' ..hehe

 

[CM_Tutor]

Originally posted by nike33
c) Hence (ill add or otherwise ) or otherwise find the locus of the orthocentre of the triangle fromed by the three tangents to the parabola drawn at P X Z

NB soln of locus is y = -a.... after all that algebra bashing it came down to this

This is a 'trick' hence question.

Note that the coordinate of D is symmetric in p, x and z.

Now, the tangents at P and X meet at K, [a(p + x), apx], and the line through K prpendicular to the tangent at Z will also meet the directrix at D - You can do all the algenra to establish this fact, but I know it's true as the coordinate of D is symmetric in p, x and z. Hence D is the orthocentre (the intersection of two of the altitudes, and the locus of D is y = -a.

 

[nike33]

yes...i can see it after i have the answer..

 

[CM_Tutor]

Another, more rigorously alegbraic approach, would be to show that the gradient of KD is -1 / z, and the result would then immediately follow.