azn_chilly's scribe post (problems + learning)

This morning we started off with three questions off the bored and i was late for that class and was who willing to the scribe today hahaha!!!!!!. so we stared off by this .....

graph this conic section and label all information points
4x(square)+9x(square)-16x+18y-11=0

4x(square)-16x + 9x(square)+18y =11
(4x(square)-16x+4-4) + (9x(square)+18y+1-1)=11+16+9
4(x-2)(square)/36 + 9(y+1)(square)/36 = 36/36
(x-2)(square)/9 + (y+1)(square)/4 =1
(x-2)(square)/3 + (y+1)(square)/2 =1

A(square) - B(square)=C(square)
3(square) - 2(square)=C(square)
9-4=C(square)
5=C(square)
(square root)5=C

find the equation of an ellipse with center at (2,-5), major axis at length 6 units parallel to the x-axis and major axis 2 units long

(x-2)(square)/9 - (y+5)(square)/1=1

a tunnel has a semi-elliptical cross-section it is 20m wide and 5m at its highest point, find the equation of the ellipse and use it to find the height of the tunnel 2m away from one end

(x-h)(square)/a(square) + (y-k)(square)/b(square)=1
x(square)/100 + y(square)/25=1

8(square)/100 + y(square)/25=1
y(square)/25=100/100-64/100
y(square)/25=36/100
(25)y(square)/25=36/100(25)
y(square)=9
y(square)-9=0
(y+3)(y-3)=0
y=-3 y=3
y=-3 is rejected cause it will be in the negative side

in the afternoon we talked about of the construction of the hyperbola and its properties...

(x-h)(square)/a(square) - (y-k)(square)/b(square)=1 is the hyperbola
(x-h)(square)/a(square) + (y-k)(square)/b(square)=1 is the ellipse

hyperbola: a locus of points that moves in such a way so that the absolute value of the distance(focal radii) from two fixed points(foci) is constant
|PF1-pf2|= 2a[a constant]

the anatomy of a hyperbola
0 is the centre
A1 + A2 are the verticies
A1A2 is the transverse axis its length is 2a
B1 +B2 are the end point of the conjugate axis its length is 2b
F1 + F2 are the foci they are for the centre
PF1 and PF2 are the focal radii
|PF1 - PF2|= 2a

the picture of the hyperbola is horizontal and centred at the the origin, the equation of the of the asymptotes are: y= b/ax and y=-b/ax
the transvers and conjugate axis are lines of symmetry for the hyperbola

Conic Scribe Post

Hey everyone it's tennyson in the house!!!!

During the morning we started to work out some problems that we had from the last homework that everybody had troubles with.

From the homework Exercise sheet

#2.

Determine the equation of the circle which passes throguht the points
J(-3,2) k(4,1) L(6,5).

Points:
J(-3,2)
k(4,1)
L(6,5)




this is an example of how we know where the centre is
( spelling error know = knowing )


we have to find the midpoint.

So we used Rise/Run

Rise/run
m(KL)= 4/2 = 2
m(AL)= -1/2

midpoint
midKe=(5,3)

y-3=-1/2(x-5)


After that the whole class has to do the paper folding again, this time for the "Hyperbola"

Here was a sample of what it turn out to be:




After that we had to calculate what line measures up to.

Pf1 - Pf2 = 56mm-135mm = 79

Qf1-Qf2 =146mm-69mm = 77

Rf1-Rf2 =143mm - 69mm = 76

A1A2= 81

( note: my calculations are different from other students)

During the afternoon mr.K was away , so we just did a quiz & homework on exercise 38

well by3 everyone , next person would be : JOHN!!!!!

ScubaSteve's Third Scribe Post

This just in! Mr.K recently informed that the Flickr assignment has been extended! Memo hurrahs around. So for those who are thinking about flying under the radar, get camera happy! Plus, the favorite number picture will still be allowed for uploading. Aside from the pc40sf06 tag, the favorite number should also be tagged with a 'interesting' and the recent trig assignment be tagged with a 'trigonometry' tag =). Enjoy.

This is ScubaSteve reporting for the third time. I know so far i've topped my post with the previous one, and hopefully you guys are not expecting me to top this one ;_;. The last one was a big one, so many pictures almost jepordized my scribe post (it was spacing out around the end, good thing i finished before it dragged on). Now, today was a double class. We were introduced to Conic Sections, a new unit that i seemed to be already confused about. Its not that i dont understand it, its just tough to re-write in my own words or explain it, so bare with me ^.^; I promise if i scribe again, i'll make up for this post if this is unsatisfactory ;]. Without further adieu, read on my cyber-math classmates.

The first class was an introduction. We learned how to find the eq'n by looking at the graph. Lets look at a basic one first.

Given what is drawn here the formula to find an eq'n of a graph is: (PF=PD) (root->),(x-h)^2+[y-(k-p)]^2 = ,(x-x)^2+[y-(k-p)]^2. First we properly expand the monster that is [y-(k-p)]^2 which ends up being:

(y-k-p)(y-k-p) => y^2-yp-yk+k^2+kp-yp+kp+p2 => y2-2yk-2yp+2kp+k^2+p^2 (its ugly i know but we'll have to bare with it) So putting it back into the original eq'n we'll have...

(x-h)^2+y^2+k^2+p^2-2yk-2yp+2kp=y^2+k^2+p^2-2yk-2yp+2kp => (x-h)^2=4yp-4kp => (x-h)^2=4p(y-k)

This graphs eq'n is (x+2)^2 = 8(y-3)

Could you recognize simularities in the eq'n that is familiarly applied to the graph? Vertex?

The p is the distance between the Focus (F) to the vertex. It is equal to the distance between the vertex and the Directrix (D)

P can be found by 4p=(coefficient), in this case it's 8, therefore 2

Back in grade 11 we learned the eq'n of a parabola was y=a(x-h)^2+b. Although this gets the job done back then, it does not provide us sufficient information when it comes to Conic Sections. y=a(x-h)^2+b can be expressed as the formula we use now, just to prove it is legitament.

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

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

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

[(x-h)^2=(1/a)(y-k)]

That ended the first period. When we came back, we had a couple of questions on the board. These questions were incomplete squares with eq'ns that were un-usuable. We had to retrace our steps back a year to balance out the eq'n before we could find any information on them.

y^2+8x-6y+1=0 find... (work in red text to find the solutions to answers)

(y is the one ^2'd so it opens sideways) (if x was ^2'd it would open up or down)

Vertex;(1,3)

Focus;(-1,3) (vertex's x value -2) <-2=p (4p is negitive, therefore it opens left, parabola is around focus)

Eq'n Of Directrix;x=3, (vertex's x +2) <-2=p (4p is negitive, therefore it opens left, parabola opens away from directrix)

Eq'n of Axis of Sym;y=2 (axis of sym = vertex's y value if open sideways, x value if up or down)

Domain;(-oo, 1] (4p is negitive, therefore it opens left)

Range;(-oo, oo)

y^2+8x-6y+1=0

y^2-6y=-8x-1

y^2-6y+'9'=-8x-1+'9'

(y-3)^2=-8(x-1)

4p=-8, p=-2

The rest of the class consisted of a math dictionary insert, enjoy!

MATH DICTIONARY IS THE FOLLOWING:

Conic Sections

The word 'conic' means 'cone-like'. Four different types of cruves are generated by taking cross sections of a double napped cone. They are: i)Circles, ii)Ellipses, iii)Parabolas, iv)Hyperbolas

Locus: Set of points that follow a certain rules

Focus: A fixed point (or points) that determines a locus

Directrix: A fixed line that determines a Locus

Parabola: A locus of points that are equidistant from a fixed point (focus) and a fixed line (directrix)

The Anatomy of a Parabola

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

vertex(h-k)

focus(h,k+p)

eq'n of Directrix y=k-p

Eq'n of axis of sym: x=h

note: p

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

vertex (h,k)

focus (h,k+p)

eq'n of Directrix: x=k-p

eq'n of axis of sym: y=k

note: p

That ends it for today's scribe post~. It's really late now, i should wrap it up. Homework is Excercise the next =o, i believe it's called 'Parabola'. Serious of not, Mr.k said test was on friday, so learn hard this week! ScubaSteve out.