Today, we went over the parabola and it's properties: these include the focus point (the point that the parabola "wraps" around), the directrix (line that is related to the position of the vertex), the vertex (coordinates h,k) and of course there is more, described throughout this post.

Slide 1:

- This is an equation of a parabola. Notice that in this equation, the exponent 2 is on the x-variable. This causes the parabola to open up or down (depending on the sign of the value p, described later)
- This is another equation of the a parabola. Notice that in this equation, the exponent 2 is on the y-variable. This causes the parabola to open left or right (depending on the sign of the value p)
- This graph represents the equation where the parabola opens up. This is due to the positive coefficient (4) of the variable "p."

Note that the distance from a point on the parabola to the directrix is equal to the distance from the same point to the focus.

Also note that there are 2 points on the same parabola that will have the same distance from the directrix and focus EXCEPT the vertex which is special.

- This graph represents the equation where the parabola opens right. This is due to the positive coefficient (4) of the variable "p."

- The vertex has the coordinates of (h, k)
- The focus has the coordinates of (h, k+p) for the parabola opening vertically, and (h+p, k) for the parabola opening horizontally.
- The directrix has the equation y=k+p opening vertically, and y=h-p opening horizontally.

Slide 2:

- This is the newly derived equation of the parabola, from yesterdays lesson, in standard form.
- This is the old equation of the parabola that we used in grade 10/11, in standard form.
- In this step, we simply rearrange the new equation.
- On the other side, we also rearrange the equation to try and find similarities between the two equations that both represent the parabola.

Remember that when "a" or the amplitude is greater than 1, we have a stretch in the parabola horizontally. When "a" or the amplitude is less than 1 but greater than 0, we have a compression of the parabola horizontally.

Now notice that since the equation is rearranged so that the coefficient is (1/a) we come to understand why this happens.

Ex: if a=2, substituting it into (1/a) gives us 1/2 which signifies a compression.

- 4p = (1/a)
- "p" is a factor in the compression or stretch horizontally of the parabola
- "p" is also the distance from the vertex to the focus and to the directrix

Slide 3:

- Sketch this equation.
- We know that the coefficient is equal to 4p so we solve for p, to find the amplitude of the parabola.
- We find that the coordinates of the vertex is (2, -1), the focus is (2, 0) and the directrix is

y = -2. Also since p is equal to 1, the amplitude is equal to 1.

- Sketch this equation.
- Find p by creating the equation -8 = 4p (since -8 is the coefficient).
- Since the exponent 2 is on the y variable, the parabola opens horizontally. Also, since the sign of p is negative, the parabola opens left instead of right.

Because the absolute value of p is greater than 1, in this case 2, the graph is stretched by a factor of 2.

We also find that the coordinates of the vertex is (3, -1), the focus is (1, -1) and the directrix has an equation of x=5. Remember to always draw the directrix with a DOTTED LINE only.

Slide 5:

- In this slide, we just sketched more examples of parabolas.
- Since John didn't have the chance to outline his sketch like Vince (Left), I shall:

This parabola opens up/down because x is squared. It opens down because the coefficient of y is negative (in this case -4)

Slide 6:

- In this step, we generalize the equation by expanding.
- After simplifying and moving every term on one side, we have the end product of the general form equation.
- After a handful of moans from the class, Mr. K attempts to show us that it is easily converted back into the standard form, first by completing the square.
- Simplifying the equation occurs.
- 4 is factored out of 4y-8.

Slide 7 Last:

- A Circle is a LOCUS (A set of points that follow a certain rule) that are equidistant from a central point. In this case, the center point "O" (h, k) is "r" (radius) away from "P" (x, y) a point on the circle.
- The length from O to P is equal to "r"

Noting this, Mr. K attempts to find this distance by using the distance formula and leads us to the end of the class with the standard form of a circle.

At last I finish my seemingly long post on my laptop. However I have good news! I just fixed my computer by replacing the network adapter (Now I can return to gaming :) ). I hope this scribe post helps whoever reads this and, I hope it is simple and easy to understand as that was and is my goal.

Hmm.. it seems like the scribe list hasn't been updated so I will choose...

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Miles! (I don't think he's been scribe yet, this cycle)

Have a great night people!

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