ANOTHER Parametric Equations of Parabola question -- need help (2 Viewers)

[blackops23]

Hi guys, here's the question:

P is a given point with parameter p on the parabola x^2 = 4ay, with focus S(0,a). A line is drawn from S, perpendicular to SP and meets the normal at P in the point Q. PN, QM are drawn perpendicular to the axis of the parabola.
(a) Find the coordinates of Q.
ANSWER: Q=(ap(1- p^2) , (a(2p^2 + 1)

Might need to sketch this one out guys
Here's my sketch on geogebra- hope its accurate

http://img42.imageshack.us/i/q14pg115jonescouchman.jpg/

On that diagram a=1, and t=3, Q=(-24,19) M=(0,19) P=(6,9) N=(0,9)
That's about it for the special points
Any help would be extremely appreciated.

Thanks guys.

 

[deterministic]

I'll give you a couple of hints and I'll let you try it by yourself.
- If 2 lines are perpendicular, their gradients multiply to -1. Let Q be (x0,y0), derive an equation with (x0,y0) using the fact QS is perpendicular to PS.
- Note that Q(x0,y0) also satisfies the equation of the normal to the parabola at P. From this, you have another equation which involves (x0, y0).

So you have 2 equations solving for 2 unknowns... guess what you do?

 

[Gussy Booo]

By letting Q be (x0,y0) we have to set ourself a goal. We have TWO UNKNOWNS.
If we somehow generate TWO equations both constituting of these TWO UNKNOWNS, they can be easily found through the process of simultaneous equations. The algebra isn't nice. Factorising involved. But deterministic has given you all the clues you need.
Good luck!

 

[blackops23]

my equation for QS turned out to be a fat piece of ****

gradient of QS: -2p/(p^2 - 1)
Using y-y1 = m(x-x1) and using the focus (0,a)
y-a = (-2px)/(p^2 -1)
therefore, y= [-2px + a(p^2 - 1)]/ (p^2-1)
Sub that into x = 2ap + ap^3 - py

and you get a real big mess
Is there another way to solve them simultaneously so it is less messy?

Thanks guys

P.S - alright no worries, I'll plow through the algebra

 

[Gussy Booo]

Hahahah! Parametrics always brings up a lot of algebra. You should see 4u conics (which is parametrics with hyperbolas etc etc), the algebra goes on and on
So, by giving Q (x0,y0) we are working towards generating two equations which will consist of Q ONLY.

1)PS in perpendicular to QS

Gradient of PS = (p²-1)/2p
Gradient of QS = (y0-a)/x0
We know the rule that two lines which are perpendicular result in the product of their gradients equaling -1.
Therefore:

[(p²-1)/2p][(y0-a)/x0] = -1
(1-p²)(y0-a)=2px0 -> EQUATION 1 (I've just done a algebra to make it look nicer.)

2) Q lies on the Normal.

Understandably any point which lies on a curve SATISFIES the equation of that curve.
We know that Q lies on the normal ap³-yp=x-2ap
Hence the substitution of Q(x0,y0) for x and y respectively will still hold the equation true
Therefore: ap³-y0p=x0-2ap -> EQUATION 2

(1-p²)(y0-a)=2px0 -> EQUATION 1
ap³-y0p=x0-2ap -> EQUATION 2

Now Algebra time.
I'm going to make x0 the subject of equation 2:

x0=ap³-y0p-2ap
x0=p(p²-y0+2a)

Putting this value of x0 into equation 1:

(1-p²)(y0-a)=2p[p(p²-y0+2a)]
(1-p²)(y0-a)=2p²(p²-y0+2a)

Making y0 the subject.

y0(1-p²+2p²)=(2a)p^4 + (3a)p²+a
y0(1+p²)=(2a)p^4 + (3a)p²+a
Now we have to factorise the RHS
y0(1+p²)=2ap²(p²+1)+a(p²+1)
y0(1+p²)=(2ap²+a)(p²+1)
y0(1+p²)=a(2p²+1)(p²+1)
y0=a(2p²+1)

You can find x0 by yourself <3 hahaha

 

[blackops23]

gradient of PS: (p^2 -1) / 2p
therefore gradient of QS: (2p) / (1 - p^2)
y-y1=m(x-x1) using (0,a) the focus

therefore:

y-a = (2p/ 1-p^2)*x
therefore, x= (y-a)/(2p/ 1-p^2)
x= [(y-a)(1- p^2)]/2p
equation of normal at P: x = 2ap + ap^3 - py
Solve simultaneously

[(y-a)(1-p^2)]/2p = 2ap + ap^3 - py

Multiply through by 2p
therefore:
y - yp^2 - a + ap^2 = 4ap^2 + 2ap^4 - 2yp^2
y + yp^2 - 3ap^2 - 2ap^4 -a = 0
y(p^2 + 1) = 3ap^2 + 2ap^4 + a
therefore, y = (3ap^2 + 2ap^4 +a)/(p^2 + 1)

LIKE, WTF?? that definitely is no where near to what the answer is ie. y= (a(2p^2 + 1)

So what on earth did I do wrong??

 

[Gussy Booo]

Look at mine ^^^^
Tell me which part of my working out you don't understand
You have to factorise that numerator.

 

[blackops23]

y0(1+p²)=(2a)p^4 + (3a)p²+a
Now we have to factorise the RHS
y0(1+p²)=2ap²(p²+1)+a(p²+1)
y0(1+p²)=(2ap²+a)(p²+1)

Holy crap lol, factorising the RHS was that one barrier stopping from proceeding, however I never ever would have thought of changing (2a)p^4 + (3a)p²+a to (2a)p^4 + 2(a)p^2 + (a)p^2 + a, and then factorising, my god it actually worked! HOW ON EARTH did you come up with that??
But anyways, mate, thanks a million for all the help you've given, still can't believe it took me a day to complete one question

Also, is it essential that you use Q=(x0,y0) instead of Q=(x,y) because x and y are variables.

Thanks mate!! POSITIVE REP FOR U COMING UP!

 

[Gussy Booo]

Yes well. It comes from experience. When you're doing 4u parametrics you learn a lot.
My best advice is to factorise whenever possible. Make every line of working out neater and neater. Try to eliminate ugly fractions immediately.
Looking at : (2a)p^4 + (3a)p²+a, you can see how I put the coeffecients of the p in brackets, so they stood out. I also put them in order of degree, so that I could clearly see that I had p^4, p². You can see that its a quatric equation, or a quadratic equation if letting p²=M or such.
What I mean is. Let p²=M:
(2a)M²+(3a)M +a
^^Simple and clear quadtratic equation. Then just factorise
It's all about experience my friend.