HSC 2015 MX1 Marathon (archive) (5 Viewers)

SammyT123 · Jul 4, 2015

HSC 2015 3U Marathon

davidgoes4wce said:
Would I be right in saying the directrix is y=-p?

Is this latex or an image ?
Also what is post 358, is it latex?

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davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

its an image (or screenshot) from Latex

SammyT123 · Jul 4, 2015

Re: HSC 2015 3U Marathon

I cannot view whatever is in post 358, did you take the screenshot and manually upload it to photobucket?

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davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

Sammy yes I did

davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

Here is another link from html if you want to see the question

<a href="http://s571.photobucket.com/user/Tran_David/media/Screen%20Shot%202015-07-04%20at%209.48.51%20pm_zps6p3eqhf9.png.html" target="_blank"><img src="http://i571.photobucket.com/albums/ss157/Tran_David/Screen%20Shot%202015-07-04%20at%209.48.51%20pm_zps6p3eqhf9.png" border="0" alt=" photo Screen Shot 2015-07-04 at 9.48.51 pm_zps6p3eqhf9.png"/></a>

InteGrand · Jul 4, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Would I be right in saying the directrix is y=-p?

$Assuming your final equation is correct (I haven't checked it much), the vertex is $(-2p,-p)$, and the parabola is concave up, so the directrix would be $p$ units below the directrix (since the focal length from your equation is $|p|=p$ (as $p>0$)). So the directrix would be the line $y= -2p$.$

InteGrand · Jul 4, 2015

Re: HSC 2015 3U Marathon

SammyT123 said:
I cannot view whatever is in post 358, did you take the screenshot and manually upload it to photobucket?

Sent from my iPhone using Tapatalk

Maybe you can't see it because you're on Tapatalk.

davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$Assuming your final equation is correct (I haven't checked it much), the vertex is $(-2p,-p)$, and the parabola is concave up, so the directrix would be $p$ units below the directrix (since the focal length from your equation is $|p|=p$ (as $p>0$)). So the directrix would be the line $y= -2p$.$

I still didn't quite understand how you got -2p.

SammyT123 · Jul 4, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Maybe you can't see it because you're on Tapatalk.

Yes , that's the problem

I can see latex on other forums , just not this one . I have contacted mods.

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davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

my thinking was for (x-h)^2=4a(y-k)

Vertex (h,k)

directrix (h,k-a)

I have that written down in my notes.

InteGrand · Jul 4, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I still didn't quite understand how you got -2p.

$In an upwards opening parabola, i.e. one of the form $(x-h)^2 = 4a(y-k)$, where $a$ is positive, the vertex is $(h,k)$, and the directrix is the horizontal like that is $a$ units below the vertex, which is $y= y_\text{vertex} -a$, i.e. $y = k -a$. In your particular question, we had $a=p$ and $k= -p$.$

davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

Oh wait I get it now (Im a slow learner )

=-p-p
=-2p

davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

No offence but if you are learning this subject in a 1-2 year time frame, there are so many traps, tricks etc. But I'm picking them up slowly as I get along.

InteGrand · Jul 4, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
my thinking was for (x-h)^2=4a(y-k)

Vertex (h,k)

directrix (h,k-a)

I have that written down in my notes.

$Yes, that is correct for $a>0$. In your question, you had $k=-p$ and $a=p$, so $k-a= -p-p= -2p$.$

InteGrand · Jul 4, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
No offence but if you are learning this subject in a 1-2 year time frame, there are so many traps, tricks etc. But I'm picking them up slowly as I get along.

Why would I be offended by this?

davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

Someone give me a head start to this question:

davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

This forum is like a learning tool....some genuine geniuses in here.

Carrotsticks · Jul 4, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Someone give me a head start to this question:

Top equation minus two times the bottom equation.

Speed6 · Jul 4, 2015

Re: HSC 2015 3U Marathon

The geniuses on this thread are always helping David, ask away and they shall be ready for a challenge.

davidgoes4wce · Jul 4, 2015

Re: HSC 2015 3U Marathon

x=sqrt (3) and y= sqrt (3)/2

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