Locus problems - need help! (1 Viewer)

[blackops23]

Hi guys, just to inform you, I've never done Locus before, (must have slept during class or something...), anyways now I'm desperately trying to catch up, but I don't really understand some aspects, all I know is you have to eliminate the parameters from a given point.

Anyways, heres the question:

Suppose that PQ is a focal cord of the parabola x^2 = 4ay.
(i) Find and describe the locus of the midpoint M of PQ.
(ii) Find and describe the locus of the intersection T of the tangents at P and Q and show that MT is always parallel to the x-axis(I think it meant y-axis, not sure though)
------------------------------------------------

Working out for (i)

M=(a(p+q) , [a(p^2 + q^2)]/2)
therefore: x=a(p+q)
x/a= p+q
therefore:
x^2/a^2 = p^2 + 2pq + q^2
pq=-1 as it is a focal chord
therefore

x^2/a^2 = p^2 + q^2 -2 ----------1
y= [a(p^2 + q^2)]/2
therfore
2y/a = p^2 + q^2
(2y/a) -2 = p^2 + q^2 - 2 -----------2

---1=---2
therefore:
x^2/a^2 = (2y/a) - 2
x^2/a^2 = a(2y - 2a)/a^2
therefore
x^2 = 2a(y-a)
x^2 - 2ay + 2a^2 = 0 IS THE LOCUS OF M.

Guys check me on this, I'm unsure so please correct me if i'm wrong.
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Now PART (ii)

Question said MT is parallel to x-axis, but I think it meant y-axis.

tangent at P: y=px-ap^2 ------3
tangent at Q" y=qx-aq^2 ------4
Solve simultaneously (3 - 4)
therefore:
T=(a(p+q), apq)
pq=-1 as PQ is a focal chord,
therefore:

T=(a(p+q), -a)
LOCUS is y=-a, i.e. the directrix, please check this guys, I'm not sure

Show MT is parallel to y-axis,

abcissa of M and T = a(p+q), therfore MT is parallel to y-axis.

Thanks guys.

 

[xV1P3R]

Both seem fine.

 

[deterministic]

(i) working out seems fine, you just need to describe what sort of curve the locus represents (eg. parabola, straight line, circle etc) and the main features of that type of curve (eg. for parabola: the focus and focal length, circle: radius and centre), as well as any other really significant aspects.

(ii) It is y axis i suspect. your working out looks fine though.

 

[xV1P3R]

Wait hang on. y = - a is parallel to the x-axis...

 

[mirakon]

^^^^^^ yes. the question makes sense to me tbh.

y=-a is a horizontal line, its the line where all values for y are '-a'. as such if its horizontal, its parallel to the x-axis

 

[blackops23]

It says MT is parallel to the axis, not y=-a

Also,

x^2 - 2ay + 2a^2 = 0 IS THE LOCUS OF M.

How do I find the focal length and the focus?
What I did was:

x^2 - 2ay + 2a^2 = 0
2ay = x^2 + 2a^2
y= (x^2)/2a + a
therefore y-intercept is (0,a) focus,
Axis: -b/2a = 0/... = 0
therefore vertex is the focus.

But I don't know how to find focus or focal length....

 

[mirakon]

lol my bad

MT is x=a(p+q) which is indeed parallel to the y-axis

 

[deterministic]

rearrange into (x-h)^2=4A(y-k) form, where (h,k) is the vertex of the parabola. Note that A doesnt necessarily refer to the "a" in the original parabola.
So put all the x on one side:
x^2=2ay-2a^2
x^2=2a(y-a)
x^2=4(a/2)(y-a) so the focal length A=a/2 while the vertex is (0,a) so the focus is (0, a+a/2=3a/2) and so on.

 

[blackops23]

Thanks deterministic,

But here's another problem where I don't quite understand the wording:
Q: The points P(2ap,ap^2) and Q(2aq, aq^2) lie on the parabola x^2 - 4ay. Tangents drawn at P and Q interesect at T. The chord PQ passes through R(0,3a). TK is perpendicular to PQ.
(i)Show that pq=-3
(ii) Find the locus of K.

Now i got (i) fairly easily, pq=-3, however my problem is with TK, So TK is perpendicular to PQ... Does K lie on PQ?? or somewhere else?

thanks guys

 

[janchanisek]

Thanks deterministic,

(ii) Find the locus of K.

Now i got (i) fairly easily, pq=-3, however my problem is with TK, So TK is perpendicular to PQ... Does K lie on PQ?? or somewhere else?

thanks guys

Yeh i'd assume K lies on PQ otherwise it can't really be done... this question is a bit fail lol they dont even specify K when introducing another point.... =P

 

[blackops23]

Someone, or somepeople, help me find the locus of K, or at least give me a few hinters =)

 

[janchanisek]

Someone, or somepeople, help me find the locus of K, or at least give me a few hinters =)

okiesss
1. Find point T by simultaneous equations of tangents at P and Q
2. Find gradient of chord PQ..... m = (p+q)/2
3. Find the equation of chord PQ .... y = (p+q)x/2 - apq ....
4. Find the equation of TK using m = -2/(p+q) and point T [(a(p+q), apq] ..... y = -2x/(p+q) + apq + 2a
5. Substitute pq = -3 as you found before.... so PQ: y = (p+q)x/2 + 3a; and TK: y = -2x/(p+q) - a
6. Rearrange both equations (4) and (5) to make the terms with x as the subject
so equation of PQ: (p+q)x/2 = y - 3a
and equation of TK: 2x/(p+q) = -y - a
7. Multiply the two equations together (LHS1*LHS2 = RHS1*RHS2) and your p and q's will disappear
so.... x^2 = (y - 3a)(-y - a)
8. Expand and make it in the form x^2 + (y-k)^2 = c

answer: x^2 + (y-a)^2 = 4a^2

DONE =D
Hope this helps! =]

 

[Drongoski]

(Sorry for this side track !!)

janchanisek

just curious. Do you eat durian at all? If so which type?

- Thai
- Vietnamese
- Malaysian
- Indonesian
- other

 

[blackops23]

.

 

[janchanisek]

(Sorry for this side track !!)

janchanisek

just curious. Do you eat durian at all? If so which type?

- Thai
- Vietnamese
- Malaysian
- Indonesian
- other

LOL yeh i like durian.... mostly eat thai i think haha
in fact im widely known as 'Durian' in certain circles lol =P
have i met a fellow durian lover?

 

[blackops23]

okiesss
1. Find point T by simultaneous equations of tangents at P and Q
2. Find gradient of chord PQ..... m = (p+q)/2
3. Find the equation of chord PQ .... y = (p+q)x/2 - apq ....
4. Find the equation of TK using m = -2/(p+q) and point T [(a(p+q), apq] ..... y = -2x/(p+q) + apq + 2a
5. Substitute pq = -3 as you found before.... so PQ: y = (p+q)x/2 + 3a; and TK: y = -2x/(p+q) - a
6. Rearrange both equations (4) and (5) to make the terms with x as the subject
so equation of PQ: (p+q)x/2 = y - 3a
and equation of TK: 2x/(p+q) = -y - a
7. Multiply the two equations together (LHS1*LHS2 = RHS1*RHS2) and your p and q's will disappear
so.... x^2 = (y - 3a)(-y - a)
8. Expand and make it in the form x^2 + (y-k)^2 = c

answer: x^2 + (y-a)^2 = 4a^2

DONE =D
Hope this helps! =]

Thanks mate, it helped a lot, however if there was a quicker method, I'd really appreciate it. because in an exam, I would be screwed, I also noticed that you multiplied the equations, WOW never come across that before, but I guess it really smoothens things in this situation.

 

[Drongoski]

LOL yeh i like durian.... mostly eat thai i think haha
in fact im widely known as 'Durian' in certain circles lol =P
have i met a fellow durian lover?

Yep - I love Malaysians mainly. Unlike the rather large Thai ones (found sold in Asian groceries) that are harvested off the tree, the Malaysian ones are more flavoursome (or stinking for those who do not like them). In Malaysia durians are picked after they have fallen off the tree.

It's amazing you enjoy durians if, presumably, you were born & bred here.