Locus problems (1 Viewer)

[smallcattle]

i dont understand this topic.. so sort of questions can they ask??

 

[velox]

Do you mean Locus and the parabola ? I.e Parametrics (x² = 4ay)?

 

[smallcattle]

Yeh, that topic, i cant find much questions on that topic in past papers, so i have a strong feeling that they might put it up this time

i just dont understand..this topic

 

[velox]

Well just learn how to derive all the equations such as, the tangents to a parabola, normal to a parabola etc. After knowing the derivations, most questions are reasonably ok. Oh wait, parametrics is 3u.

So you'd be talking about the locus of circles etc...

 

[Seraph]

look at teh 2002 paper

Co-ordiante geometry question (iii)

Find the co-ordinates of poitn C that lies on the y axis and is equidistiant from A and B

now imagine this graph plz lol B(1,5) and A(2,2)

now in this case essentially we aer trying to find point C , it has been made very easy for us because they have told us its on the y axis this means x is 0

so using locus we can find the poitn

the term Equidistant means essentially The distance from C to B is equal to distnace from C to A
you should draw this , C is (0,y) now our condition is then
CB = CA

very good, now that being said we use our distance formulas

((5-y)^2 + (1-0)^2)^(1/2) = ((2-y)^2 + (2-0)^2)^(1/2)

very good
now , we get rid of the square roots , by x each side by them

and just make the equation y = something
adn eventually we get our Y point and there we have point C

this is just a typical application of LOCUS

 

[smallcattle]

locus of the parabola, focal length, directrix and things like that

 

[Seraph]

?

(X-h)^2 = 4a(y-k) is your equation for a parabola that can be anywhere but in a direction where its y axis is its symmetry

(h,k) vertex
4a its positive we know its direction is upwords

HERE IS A TYPICAL QUESTION

Consider the parabola y = x^2 - 8x + 4

find its vertext
we must change it into that equation

TO DO SO WE NEED TO COMPLETE ZE SQUARE

x^2 - 8x = y -4
x^2 -8x + (-8/2)^2 = -4 + (-8/2)^2 complete ze square is just square the -b thing and divide by 2

so x^2 -8x + 16 = y - 4 + 16
x^2 - 8x + 16 = y + 12
now simplify to the (X-h)^2 = 4a(y-k)

(x-4)^2 = Now because this equation is gay on this side we cant really factorise anythign so its form is
(x-4)^2 = (y +12)
however we must find our FOCAL LENGTH
this is a positive parabola because in front of (y+12) its a 1
so 4a = 1
a = 1/4
ORU FOCAL LENGTH IS 1/4
NICE
so ... our vertex is just h.k right?
4, - 12 IS OUR VERTEX
great
OUR FOCUS is then (since its concave up) we add 1/4 TO THE Y POINT , remember this is a positive PARABOLA
so our focus co-ordiantes are (4,-11 3/4)


P.s The directrix is just The opposite direction of the focal length to the Vertex Point
so in this case.... it would be y = -12 -1/4
y = -12.25

similarly if it was -4a we would work in the opposite direction
and if it was (y-k)^2 y is dominant this means the parabola is flipped along the x axis..... manipulation thats all it is

 

[smallcattle]

very nice explaination.. very understandable lol.. i'll rep u for that

 

[Seraph]

hah if i can find one

but i think the locus of the circle questions are the same principle

only you will get an equation and it will have x^2 and y^2

so you manipulate it into the circle forumla

x^2 + y^2 = r^2
or (x-h)^2 + (y-k)^2 = r^2 for a circle at any place (hah im pretty sure)

oh and you guys should know about the perpendicular locus

with these questions your condition is

GRADIENT PA x GRADIENT PB = -1

you must use the gradient formula's and manipulate it using this equation
sorry i'll try to find some questions Right after i do some practice on shit superannuation
NOTE I AM YET TO SEE THIS QUESTION IN ANY HSC PAPER , BUT KNOWING THE BOS MARKERS THEY ARE PROBABLY LOOKING AT THIS POST AND THINKING LOL OMG THAT SERAPH WHAT A N00B LETS THROW ONE IN

no seriously though , locus is not that common , but yea i guess its good not to take any chances

 

[Binky]

?

(X-h)^2 = 4a(y-k) is your equation for a parabola that can be anywhere but in a direction where its y axis is its symmetry

(h,k) vertex
4a its positive we know its direction is upwords

HERE IS A TYPICAL QUESTION

Consider the parabola y = x^2 - 8x + 4

find its vertext
we must change it into that equation

TO DO SO WE NEED TO COMPLETE ZE SQUARE

x^2 - 8x = y -4
x^2 -8x + (-8/2)^2 = -4 + (-8/2)^2 complete ze square is just square the -b thing and divide by 2

so x^2 -8x + 16 = y - 4 + 16
x^2 - 8x + 16 = y + 12
now simplify to the (X-h)^2 = 4a(y-k)

(x-4)^2 = Now because this equation is gay on this side we cant really factorise anythign so its form is
(x-4)^2 = (y +12)
however we must find our FOCAL LENGTH
this is a positive parabola because in front of (y+12) its a 1
so 4a = 1
a = 1/4
ORU FOCAL LENGTH IS 1/4
NICE
so ... our vertex is just h.k right?
4, - 12 IS OUR VERTEX
great
OUR FOCUS is then (since its concave up) we add 1/4 TO THE Y POINT , remember this is a positive PARABOLA
so our focus co-ordiantes are (4,-11 3/4)


P.s The directrix is just The opposite direction of the focal length to the Vertex Point
so in this case.... it would be y = -12 -1/4
y = -12.25

similarly if it was -4a we would work in the opposite direction
and if it was (y-k)^2 y is dominant this means the parabola is flipped along the x axis..... manipulation thats all it is

Just thought i'd point out here you don't actually need to complete the square...
i find this way easier, so have a think about it...
ok first find the axis of symmetry with: -b/2a
then sub that y value in and get its x value and then you have your coordinates of the vertex, then just stick in in: (x-h)^2 = 4a(y-k) then you can work out what a is.