Post by: dermite on January 28, 2018, 09:53:22 am
I'm in need of some parametrics help (pls look at attached qn), as i really don't understand the topic. I'm also looking for ways to approach these types of qns.
Thanks
Post by: Opengangs on January 28, 2018, 10:06:58 am
So:
Edit: whoops, didn't realise they gave you the equation.
From the equation, just substitute (0, a) to show that pq = -1 since a focal chord is defined to satisfy the chord and passing through the focus (0, a).
Also, here are some tips to parametric equations:
FOCAL CHORD: Find the equation of the chord first, then substitute the focus into the equation.
DIRECTRIX: Know that it's defined as: y = -a for x^2 = 4ay and x = -a for y^2 = 4ax
LOCUS: Your goal is to eliminate the parameter (p, q, t) to have it in terms of x and y. It's essentially a combination of its Cartesian equation.
Post by: dermite on January 28, 2018, 12:12:50 pm
(b)
xP=8a
∴2ap=8a⇒p=4
How did you get to the 2nd line from the first?
soz if this is a dumb qn, im really bad at parametrics
Post by: RuiAce on January 28, 2018, 12:16:52 pm
Sorry, I dont understand how you did part (b).
(b)
xP=8a
∴2ap=8a⇒p=4
How did you get to the 2nd line from the first?
soz if this is a dumb qn, im really bad at parametrics
Note that if he considered the \(y\)-coordinates, he would arrive at \( ap^2 = 16a\). But he chose not to, because it was not necessary for this question.
Post by: dermite on January 28, 2018, 12:22:05 pm
Didnt really understand how p = -1/q. Is this a formula?
Post by: RuiAce on January 28, 2018, 12:23:02 pm
Didnt really understand how p = -1/q. Is this a formula?
Post by: Opengangs on January 28, 2018, 01:19:43 pm
Didnt really understand how p = -1/q. Is this a formula?Hey, sorry about the pretty bad explanation. Was rushing it a bit!
So, we are given two pieces of information about the point P. We know it is parametrically defined by the coordinates (2ap, ap^2). We are also given that it's given by the coordinates (8a, 16a). So by simply observing this, we can deduce that 2ap must be the same as 8a (in essence: 2ap = 8a since they are essentially the same point). This gives us a value for p being 4.
Now, from part (a), we've shown that pq = -1 iff PQ is a focal chord. Since it is, then rearranging the equation, we get: q = -1/p, where p =/= 0. Then, q = -1/4 since p = 4.
Hopefully, this clears things up a bit. Again, sorry for the rushed explanations!!
Post by: dermite on January 29, 2018, 09:30:50 am
Post by: RuiAce on January 29, 2018, 09:39:09 am
I've come across another question from 2010 that doesn't have any parts, so how do I approach this?
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Alternatively, we could prove that the diagonals bisect each other at right angles. But that's WAY too much effort since we have no previous parts.
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And of course, at that point the question is complete. But I will make a note as to why.
Explaining PS=PM
Back when you were first taught 2U locus, you defined the parabola to be this:
So what you did was that you'd set \(S\) to be the point \( (0,a) \), and the line to be \( x = -a\). You would then do something like this:
\begin{align*}PS &= PM\\ PS^2 &= PM^2\\ (x-0)^2 + (y-a)^2&= (x-x)^2 + (y+a)^2\end{align*}
..and so on. What happened, was that once you tidied the mess up, you arrived at \(x^2=4ay\).
The point, however, is that \( \boxed{PS=PM} \) was the starting point to begin with. That thing underlies EVERYTHING we've done in 2U locus and 3U parametrics. It is called the "focus-directrix" definition of the parabola, and is so easily forgotten because it was our foundation that we just kept building on and eventually neglected.
So what you did was that you'd set \(S\) to be the point \( (0,a) \), and the line to be \( x = -a\). You would then do something like this:
\begin{align*}PS &= PM\\ PS^2 &= PM^2\\ (x-0)^2 + (y-a)^2&= (x-x)^2 + (y+a)^2\end{align*}
..and so on. What happened, was that once you tidied the mess up, you arrived at \(x^2=4ay\).
The point, however, is that \( \boxed{PS=PM} \) was the starting point to begin with. That thing underlies EVERYTHING we've done in 2U locus and 3U parametrics. It is called the "focus-directrix" definition of the parabola, and is so easily forgotten because it was our foundation that we just kept building on and eventually neglected.
Post by: dermite on January 29, 2018, 09:55:39 am
____________________________________
And of course, at that point the question is complete. But I will make a note as to why.Explaining PS=PMBack when you were first taught 2U locus, you defined the parabola to be this:
So what you did was that you'd set \(S\) to be the point \( (0,a) \), and the line to be \( x = -a\). You would then do something like this:
\begin{align*}PS &= PM\\ PS^2 &= PM^2\\ (x-0)^2 + (y-a)^2&= (x-x)^2 + (y+a)^2\end{align*}
..and so on. What happened, was that once you tidied the mess up, you arrived at \(x^2=4ay\).
The point, however, is that \( \boxed{PS=PM} \) was the starting point to begin with. That thing underlies EVERYTHING we've done in 2U locus and 3U parametrics. It is called the "focus-directrix" definition of the parabola, and is so easily forgotten because it was our foundation that we just kept building on and eventually neglected.
Alternatively, cant we use the distance formula to show PS = PM? We know the coordinates of P, S and M.
Also, do we have to state why PS=PM?
Post by: RuiAce on January 29, 2018, 09:58:14 am
Alternatively, cant we use the distance formula to show PS = PM? We know the coordinates of P, S and M.Yes. The long way is just to use the distance formula.
Also, do we have to state why PS=PM?
If you go by the distance formula then your reasoning is obvious. But if you're using the focus-directrix definition, you will need to state that.