My Math Forum Find "Focus and Control Line of Parable"

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 May 18th, 2015, 06:53 AM #1 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 Find "Focus and Control Line of Parable" So I have this equation in my math book. $\displaystyle (x^2)/2$ This is the "orginal" formula for a parable. Focus: (0, a) Control line: y = -a $\displaystyle (x^2)/4a$ They want me to find the focus and the control line of the parable. I have found two coordinates that the parable will hit and that is: ( 2, 2 ) and ( 4, 8 ) But I have no idea of how to find the focus and control line . . . Thanks for help!
May 18th, 2015, 07:36 AM   #2
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 Originally Posted by DecoratorFawn82 So I have this equation in my math book. (x^2)/2 $\displaystyle \ \ \$y = (x^2)/2 <----- This is an equation. This is the "orginal" formula for a parable. $\displaystyle \ \ \$It's spelled "parabola." Focus: (0, a) Control line: y = -a $\displaystyle \ \ \ \$What you call a "control line" is typically called a "directrix." (x^2)/4a $\displaystyle \ \ \ \$ This should be an equation and written, for example, as y = (x^2)/(4a). They want me to find the focus and the control line of the parable. I have found two coordinates that the parable [parabola] will hit and that is: ( 2, 2 ) and ( 4, 8 ) $\displaystyle \ \ \ \$Those points are on there, but you don't need them for this problem. But I have no idea of how to find the focus and control line . . . Thanks for help!
The vertex is always halfway between the focus and the directrix.

y = (x^2)/2 = (x^2)/(4a)

So set 2 = 4a and solve for a:

2 = 4a

1/2 = a

You stated your focus is (0, a) and your control line (directrix) is y = -a.

Now make the substitutions with the a-value.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Here are some Internet sites that discuss the vertex, focus, and directrix
(what you called a "control line") for a parabola:

.

Last edited by Math Message Board tutor; May 18th, 2015 at 07:40 AM.

 May 18th, 2015, 07:58 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 y = (x^2)/(4a) y = x^2 / (4a) is sufficient.
 May 18th, 2015, 08:13 AM #4 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 I'm sorry about that I use wrong words for some mathematical things. But it's a huge difference between what we call these things in Swedish and in English. But thanks for help, that was the answer I was looking for EDIT: How would I find the focus if I weren't given it (Focus: (0, a)) or is this the formula that is right for all parabolas? Or how would I otherwise do? Last edited by DecoratorFawn82; May 18th, 2015 at 08:36 AM.
May 18th, 2015, 10:03 AM   #5
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Quote:
 Originally Posted by Math Message Board tutor y = (x^2)/2
I am officially "scolding" you for your ignorance.
You are confusing students with redundant brackets.
then this is required: y = (x^2)/(2).

I'll be reporting this post of yours
to the STD (Suspected Trolls Department).

You're in trouble deep: the STD CEO
is none other than Sir Jonah The Guzzler

 May 19th, 2015, 04:09 AM #6 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 What?! I try to explain as well as I can. And I try to write the words correct words in mathematics in English. I'm writing as well as I can and I try to explain what I mean. I don't really get why you would report my posts? I'm only asking for help because I don't know how to solve these problems on my own. And no I'm not trying to troll someone. My posts are serious and I will try to do better in my explanations then. I'll do as good as I can.
 May 19th, 2015, 05:15 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 That was NOT addressed to you...and is only a joke...OK? You're doing quite good for someone learning English.
 May 19th, 2015, 05:29 AM #8 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 I didn't see that at first. Okay. Thank you then!

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