locus!! help!! (1 Viewer)

[da_butterfree]

hey!! can anyone help with locas??? its the worst topic!! any easy ways to understand or remember it?? any tips?? please help!!

 

[Winston]

What particular part of Locus? :S...

 

[KeypadSDM]

The algebra part

 

[Affinity]

yeah... go back and do chapters 1-3.

 

[~*HSC 4 life*~]

Originally posted by da_butterfree
hey!! can anyone help with locas??? its the worst topic!! any easy ways to understand or remember it?? any tips?? please help!!

always draw a diagram!

 

[da_butterfree]

locus of the parabola.. the one where it you put in the h and the a and f and theres a big formula.. something that talks about finding things like focus or directrix when teh parabola is facing different positions!!

 

[~*HSC 4 life*~]

What don't you understand about it? find a good textbook that goes through the theory of it really nicely such as maths in focus- it has good explanations...and then try some q's..if you get stuck ask your teacher or come here again

 

[da_butterfree]

i did go thru the text and look at other books too.. what i dont get is the wiered formulas or rather say "equations" using h and k!! like when the parabola is away from the origin, u change stuff in x2= 4ay thingy. by adding h and k and that changes the focus and directrix and i just cant seem to solve sums that talk abt stuff like that!@@

 

[Heinz]

Originally posted by da_butterfree
i did go thru the text and look at other books too.. what i dont get is the wiered formulas or rather say "equations" using h and k!! like when the parabola is away from the origin, u change stuff in x2= 4ay thingy. by adding h and k and that changes the focus and directrix and i just cant seem to solve sums that talk abt stuff like that!@@

It isnt that much different then when the focus is at the origin. Draw the picture if you get stuck?

 

[Winston]

All you really need to know is this

x^2 = 4ay

this equation is the equation where the vertex is at the origin (0,0)
the focus is (0,a)
the directrix as y = - a

But for a parabola where it's vertex is not centre origin (0,0) the equation is

(x-h)^2 = 4a(y-k)

that's the equation of a Parabola concave up with the vertex being (h,k)

With the focus being (h,k+a)
With the directrix as y = k-a

Here's the complete table for you to remember, personally it's a personal desire if you want to remember them fine... it will be quick in some terms in the actual exam... but sometimes they might ask you to sketch it and plot all the vertex, focus, and the directrix.

So you should also learn about the basic sketch of it and how do draw it, as suggested from the aforementioned posts, find a good textbook, look between a few and see which explanation explains it a bit better.

Also remember one thing which is quite helpful when you're drawing it and you only have like say for example two pieces of data... say you had the directrix and the vertex, one fact that we know is : From the Focus to the vertex, this n number of units, is exactly the same length as from the vertex to the directrix, so here below is an illustrated example of what i'm saying:

Here are two additional sites i've googled that you should have a read of:

http://www.tpub.com/math2/13.htm

http://colalg.math.csusb.edu/~devel/precalcdemo/conics/src/parabola.html

Hope this helps you understand it a little bit better

 

[Winston]

Attachment for above posts.

 

[Grey Council]

whoosht! Huy the second! lol, nice post Winston.

 

[da_butterfree]

how do u find the equation of the parabola when u are given with a focus and directrix??

 

[jesshika]

Originally posted by da_butterfree
how do u find the equation of the parabola when u are given with a focus and directrix??

u would have to find the vertex and "a" if it's not at the origin and sub into the equation that winston has written up dere

" (x - h)^2 = 4a (y - k) "

kekek that didn't answer your q ... so heerez an example i found
in my notes

1. parabola has the line y=-3 as its directrix and the point (0,1) as its focus ...

Find the equation of the parabola

1st : Find 'a' .....
2nd: Find the vertex
3rd: sub into eqn formula

FINdING 'a'
The distance between the focus and directrix is 2a .. so
therefore :
....... 2a = 1 + 3 (forget the negative infront of the 3)
....... a = 2

VERTEX
can be found by subtracting 'a' from the focus (y value) {focal length }
... i.e. 1 - 2 = -1
so vertex is (0, -1)

so den you juss sub the vertex and 'a' into that formulaeee..

so it becomes

(x-0)^2 = 4(2) (y - -1)

x^2 = 8(y+1)