Saturday, May 10, 2008

Parabolas and More...

What we did in todays class:

1) learned formula of a horizontal and vertical parabola
2) graphing the focus, directrix, vertex and I almost forgot the parabola
3) completing the square

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




From that piece of paper where we did all those folds, we realized that the distance of PF was equal to the distance of PD. So by putting the coordinates of PF and PD into the distance formula we can get the formula for vertical parabolas. As you can see from the slide we first expanded (y-(k+p))^2 and (y-(k-p))^2. Also (x-x)^2 = 0 if you were wondering where that went. After expanding, we see that most of the like terms cancel out except the py's and the kp's. Then after isolating (x-h)^2 and factoring out the 4p from the right side, we're left with (x-h)^2 = 4p(y-k). The equation for the vertical parabola. The same process is done with finding the horizontal parabola's equation except the coordinates are different.


y=a(x-h)^2+k vs. (x-h)^2=4p(y-k)

-the binomial (x-h) is squared on both
-by moving k to the left side of first equation we'll see that 1/a is equal to 4p
-if a>1 we use to say that the parabola would be narrow, however in our new equation when 4p >1 then it will be wide

From looking at the equation we know:
-vertical parabola since equation starts of as (x-2)^2
-vertex coordinates are (2,-1)
-direction of parabolas opening, which is up
-wider than a standard parabola

To find the focus and directrix, we must find the value of p. p is the distance of the focus to the vertex and the distance between the vertex and the directrix. From the equation we can see that 4p=4 and we get p =1. Now we know the focus is 1 unit above the vertex and the directrix is down 1 unit of the vertex. Then you can find the roots by making y = 0 and voila there's our forgotten parabola.

(y+1)^2 = -8(x-3)...equation tells us:

-horizontal parabola since it's (y+1)^2
-4p is negative so opens to the left-parabola going to be very wide because 4p > 1

-v=(3,-1)

*note* kinda tricky because x and h aren't in the same position as last question

Just like last question, we should find p first. p = -2 since we know the -8 = 4p. We'll ignore the negative sign since distance is always positive. So now we know the the directrix is 2 units to the right of the vertex and the focus is 2 units to the left of the vertex.


Completing the Square!


Feels like grade 11 all over again. Anyways, Mr. K said that most of the time equations will be given to us in general form. General form is when all the terms are on one side of the equation. So to get it back into standard form we complete the square. First we get all the like terms to one side. We usually put the variable that is squared along with its like terms to the left so it'll match the equations for horizontal/vertical parabolas. So to complete the square we take the second terms coefficient and divide it by 2 then square it. Now what we add to the left side of the equation we must do to the right. Then we factor the left side and factor out the -8 from the right and we're back to standard form.

Well there is my short and early scribe post. You better be reading this Nelsa *squints*. Anyways, next scribe is.....who to chooose?! Wait there's only Kristina left psha i want more choices jk. Yeah so kristina your next.

2 comments:

paAni said...

why we must use 4p for parabola equation?

dkuropatwa said...

Look here. Look at slide #4 very closely.