Thread: Standard Form of Parabola

Hi, I've recently just learned about Quadratic Function. On my previous homework there was a question which asked me to write in standard form the equation of a parabola. The given information was...

Focus (-2, 0); directrix x = 4

I didn't and still don't understand how to solve this problem. I think I'm supposed to use the standard form of equation for a parabola,
y = a(x - h)^2+k, but I'm not quite sure how I should go about doing it. Can someone please explain this to me? Thank you.

The Vertex, a point on the parabola, is exactly half way between the focus and the directrix. These three features, focus, directrixm and vertex, should be sufficient to orient your parabola.

In particular, "y = something" as the directrix suggests the standard form you have written. If it is "x = something", you must switch it around a bit. There is not just one standard form for a parabola.

"I'm not quite sure how I should go about doing it" -- Give it a try after locating the vertex.

Sorry, but I really don't know anything about this topic. How would I find the vertex in this case? On the chart about parabolas, it says something like the vertex is (h,k)? And it also gives me a list of the focus (h, k+ 1/4a) and directrix (y=k-1/4a). What are they suppose to represent?

Did you read the section?
Do you just have a bag of formulas with no directions?

Really, if you know that little about it, you will not be benefitted much from this discussion.

Show that you can do something and maybe we can continue. Find the Vertex. You have the Focus. You have the Directrix. You have the information that the Vertex is exactly between them (equal distance). That is enough information.

Originally Posted by Unknown

Sorry, but I really don't know anything about this topic. How would I find the vertex in this case? On the chart about parabolas, it says something like the vertex is (h,k)? And it also gives me a list of the focus (h, k+ 1/4a) and directrix (y=k-1/4a). What are they suppose to represent?

Hint: The focus is on the x-axis; the vertex is on the x-axis. As TK pointed out, the vertex is half way between the focus and where the directrix crosses the x-axis. The x coordinate of the focus is -2. The x coordinate of where the directrix crosses the x-axis is 4. What is the x coordinate of the half-way point? This will give you h of the vertex (h, k). k will be 0 since the point lies on the x-axis. Vertex is at (h, 0).

Also, if you sketched the graph, you would see that the parabola opens to the left. The form of the equation you need is:

$\displaystyle \boxed{x=a(y-k)^2+h}$ where $\displaystyle a<0$

And, the focus and directrix is not what you stated. It should be:

Focus$\displaystyle \left(h+\frac{1}{4a}, k\right)$

Directrix: $\displaystyle x=h-\frac{1}{4a}$

You need to use the above information to solve for a. And remember a will be negative.

Originally Posted by TKHunny

Did you read the section?
Do you just have a bag of formulas with no directions?

Really, if you know that little about it, you will not be benefitted much from this discussion.

Show that you can do something and maybe we can continue. Find the Vertex. You have the Focus. You have the Directrix. You have the information that the Vertex is exactly between them (equal distance). That is enough information.

Yes, TKHunny, I seriously am that clueless on this topic. In fact, everything is all new to me. I was just taught this lesson in 2 hours, but I really don't get it. And you're right...this discussion probably won't benefit me much. So I think I'll just wait until I learn it in my school. At least by then I'll be able to learn it step by step without having to rush through it. And thank you for your explanation, masters. Your help is greatly appreciated.

Originally Posted by Unknown

Yes, TKHunny, I seriously am that clueless on this topic. In fact, everything is all new to me. I was just taught this lesson in 2 hours, but I really don't get it. And you're right...this discussion probably won't benefit me much. So I think I'll just wait until I learn it in my school. At least by then I'll be able to learn it step by step without having to rush through it. And thank you for your explanation, masters. Your help is greatly appreciated.

Good luck and don't hesitate to post other problems you may encounter. We may not have time to teach things in depth, but we do our best to get you started in the right direction.....and maybe ease some frustrations along the way.

Originally Posted by Unknown

Yes, TKHunny, I seriously am that clueless on this topic. In fact, everything is all new to me. I was just taught this lesson in 2 hours, but I really don't get it. And you're right...this discussion probably won't benefit me much. So I think I'll just wait until I learn it in my school. At least by then I'll be able to learn it step by step without having to rush through it. And thank you for your explanation, masters. Your help is greatly appreciated.

I still think you could have found the vertex.