Parabola's with Focus and Directrix

[Greed]

Hi,
Here's the problem: 2x^2=-y I'm supposed to find the Focus and Directrix of the parabola. I'll try it on my own, but I was just wondering if some1 could verify what I was doing correctly and point out what is wrong in my work.

---> has no true mathmatical value. It's just my way of saying "moving on" or just trying to show that my train of thought leads me to whatever follows that arrow (--->)

1st- I need to isolate x, correct? (2x^2)/2= x^2 which means that I have to divide the other side by 2 as well...
-y/2---> leaving the new solution to be: x^2=-y/2

2nd- Now, I need to find p? so to get p, you have to find what number times 4 equals whatever number y is corresponding with? in this case that would be 1/2 right? {I say this because -y/2 is equivalent to (-y)(1/2)} so then I need to divide 1/2 by 4, or multiply 1/2 times 1/4... which equals 1/8----> this is p?

3rd- I plug "p" (4)(1/8) into the equation---> x^2=-y(4)(1/8) I'm not sure if it should instead be written as---> x^2=(4)(1/8)(-y) that might be more precise, please advise me as to which form to use.

4th- The formula is finally in Standard Form!! ---> x^2=(4)(1/8)(-y)

5th- So... the Focus= (h,k+p) A friend of mine said that h essentially means x, and k is the same as y, which at this point both equal 0 I believe. (Please explain how x^2 can equal 0 in this kind of equation. Is it like a partial quadratic formula where x^2=0 and thus x=0 as well? for
-y equal to 0, I'm really confused because it's negative.) So, I plug 0 in for h, and k and I'm left with adding p (which we have already found to be 1/8) so the Focus should be (0,1/8)

6th- The Directrix: (y=k-p) so, k=0 from earlier findings, and p=1/8 ---> Directrix should be: (y=0-1/8) ---> (y=-/1/8)

Sorry if I seem like a complete idiot for posting all this information on one thread, but it did seem to help me figure out the problem as I tried to describe it in words.... that's assuming that my processes were correct. Please, if anything is wrong, tell me and explain what I should do instead. I'm not looking for an answer to the problem, rather a way of solving these problems so that I can do well on my test next week.

 

[soroban]

Hello, Greed!

You forgot the "minus", but the rest is excellent!

Don't apologize for showing your work and your reasoning.
It was great . . . thank you!

\(\displaystyle 2x^2\:=\:-y\;\;\)Find the Focus and Directrix of the parabola.

With x-squared, it is a "vertical" parabola (opens up or down).
\(\displaystyle \;\;\)General form: \(\displaystyle \,(x\,-\,h)^2\:=\:4p(y\,-\,k)\)

You rewrote it: \(\displaystyle \,x^2\:=\:-\frac{1}{2}y\)

We see that: \(\displaystyle \,h\,=\,0,\;k\,=\,0,\;p\,=\,-\frac{1}{8}\)

The vertex is \(\displaystyle \,(0,0)\,\) and it opens downward: \(\displaystyle \,\cap\)


Focus: \(\displaystyle \,(h,\,k+p)\;=\;\left(0,\,0-\frac{1}{8})\right\;=\;\left(0,\,-\frac{1}{8}\right)\)

Directrix: \(\displaystyle \,y\:=\:k-p\;\;\Rightarrow\;\;y\:=\:0\,-\left(-\frac{1}{8}\right)\;\;\Rightarrow\;\;y\:=\:\frac{1}{8}\)

                    |
            - - - - + - - - -  D: y = 1/8
                    |
                    |
      -------------***-------------
              *     |     *
           *        |        *
         *         F*(0,-1/8)  *
                    |
       *            |            *
                    |

 

[Greed]

Ok, thank you. Yah, I looked at it again and noticed that I forgot the negative. Thanks for showing the graph of it though, very helpful