Parabolas!

I am going to be showing how to do something we’ve learned in the past week. It can be hard at first but after you get the hang of it, it'll be really easy. We’re going to be working on solving Parabolas. There are 5 simple steps to working these types of problems. The only complicated part is that you do everything opposite as what you would think.

The first thing I’m going to show you is the 5 steps:

Step 1: The first step is going to be to find out what direction the Parabola is going to be facing. If you have an x^2 it will be facing up, if you have a –x^2 it will be facing down, if you have a y^2, it will be facing towards the positive side of the x-axis, and if you have –y^2, it will be facing towards the negative side of the x-axis

Step 2: The second step is to find the axis of symmetry. You do this by using the formula x=-b/2a. This will also be used to find the vertex.

Step 3: The third step is to find the Vertex. You do this by using the formula ((-b/2a, F(-b/2a)). Here’s were it can get confusing. If you have a x^2 term you will find the x coordinate, then you will plug that number into the equation. This is the equation shown above. If you have a y^2 term, you would do the same thing except you would switch the x and y coordinates.

Step 4: The fourth step is to find the focus. You do this by using the formula (1/4p)=leading coefficient. Once you find p, you use the formula, Vertex coordinate+P=Focus. This is where it can get tricky. Your vertex coordinate is probably opposite of what you are thinking. If you have an x^2 term, you will use the y coordinate and if you have a y^2 term, you will use the x coordinate. (this is written in point form)(Something that helps me remember which one to use is if there is an x^2 term, the only thing you do with the x’s is carry them down through the steps and use the y’s for everything else. If you have a y^2 term, the only thing you do is carry the y coordinates down through the steps and everything else is done using the x’s)

Step 5: The last step is going to be to find the Directrix. Since you already found p in the previous step, the is very simple. All you do is take your Vertex Coordinate and subtract p using the formula, vertex coordinate-p=directrix. (make sure this is in equation form).

I will work an example for you using the same steps as written above.

Example: Solve the following Parabola, y=(1/6)x^2

Step 1: This will be a parabola opening upwards.

Step 2: Using the equation x=-b/2a, x=-0/2(1/6) This gives you x=0, therefore, your axis of symmetry is x=0

Step 3: Because you have a x^2 term you will use ((-b/2a, F(-b/2a)). Your x-coordinate is going to be the same as your axis of symmetry. So you know that your Vertex is now (0, F(-b/2a)). Now you will plug 0 into your original equation and get y=(1/6)(0)^2 and get y=0. You now know that your vertex is (0,0)

Step 4: You automatically know that your x term is going to be 0 because it never changes when going through the steps. Now you have 1/4p=leading coeff. 1/4p=1/6, so p=(3/2) now you use the formula vertex coordinate+p= focus, so 0+(3/2)=3/2. This means that your focus is (0,(3/2))

Step 5: Since you already have P, you just use the formula Vertex Coordinate-p=directrix, so 0-(3/2)=(-3/2) This means that your focus is y=-3/2

I hope you learned something!

--Carley(: