K.Binks
  • K.Binks
How do you graph the focus (0,8) of a parabola?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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dessyj1
  • dessyj1
Hi.
dessyj1
  • dessyj1
What do you mean when you say "focus"? Are you talking about the vertex, or the y-intercept?
K.Binks
  • K.Binks
I don't really know, I had to graph the parabola y=1/32 x^2 and find the directrix and focus. I know the Directrix is 8 and someone told me the focus was (0,8) but if they hadn't, I wouldn't know how to find it or what it is on my own.

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K.Binks
  • K.Binks
they didnt explain, they just told me what it was (0,8)
dessyj1
  • dessyj1
Yeah, I do not know how to do that. Maybe @triciaal can help?
triciaal
  • triciaal
another reason not to accept answers without understanding I was thinking "doesn't this person know that the focus is just a point"
K.Binks
  • K.Binks
No, I didn't know that, which is why I was asking someone to explain
dessyj1
  • dessyj1
Sorry about that. She was talking to me.
triciaal
  • triciaal
The focus of a parabola is the point at the center of the parabola. all points on the parabola are the same distance from this point
triciaal
  • triciaal
|dw:1438625611026:dw|
triciaal
  • triciaal
the focus to the vertex is the same distance of the vertex to the directrix
triciaal
  • triciaal
1 Attachment
anonymous
  • anonymous
@K.Binks do you understand?
triciaal
  • triciaal
y = x^2 has the vertex at (0,0) the minimum point on the parabola when a is positive when a is a fraction as in less than 1 it stretches out the graph the y value is affected if a was a whole number greater than 1 it would make the graph steeper for a given x value the y value would be much greater
triciaal
  • triciaal
don't want to overload you are you understanding this so far?
K.Binks
  • K.Binks
Yes, I get that so far
K.Binks
  • K.Binks
If ya'll have more explanations, can you message it to me please? This question is taking too long, I gotta move on. Thank you
anonymous
  • anonymous
Sorry I am at work so I am in and out. Your equation to me is in the form \( \huge x^2 = 4ay \) Now to find the focus we need to find a. \( \huge y = \frac{1}{3}x^2 \) \( \huge \frac{1}{3}x^2 = 1y \) The Focus \( \huge 4a = 1\) \( \huge \frac{4a}{4} = \frac{1}{4}\) \( \huge a= \frac{1}{4}\) The directrix = x = 1/4 The latus rectum \( \huge x^2 = 1*\frac{1}{4} \) \( \huge \sqrt{x^2} = \sqrt{\frac{1}{4}} \) \( \huge \sqrt{x^2} = \sqrt{\frac{1}{4}} \) \( \huge x = \frac{1}{2} \) So the latus rectum = \(\huge (\frac{1}{2}, \frac{1}{4}) \) and \(\huge (-\frac{1}{2}, \frac{1}{4}) \) Now we have a stretch though. I have to leave to go to the gym but when I go back home I will finish up and explain more. Here is the graph, which I am not done.
anonymous
  • anonymous
https://www.desmos.com/calculator/css6cjlswn
anonymous
  • anonymous
The directrix = y = -1/4
anonymous
  • anonymous
Anyways, from my post we have the focus, directirx and the latus rectum from the equation \( \huge y = x^2\ \) which is \( \huge x^2 = 4ay \) They only different in yours \(\huge y= \frac{1}{3}x^2 \) is that you have a stretch. Well, since we have all our points, all we need to do is stretch the points so Focus = \( \huge (0,\frac{1}{4}) \) Latus Rectum = \( \huge (\frac{1}{2},\frac{1}{4}) \) and \( \huge (-\frac{1}{2},\frac{1}{4})\) Now we just need to stretch them so after the stretch Focus = \( \huge (0,\frac{1}{3}) \) Latus Rectum = \( \huge (1,\frac{1}{3}) \) and \( \huge (-1,\frac{1}{3})\) And we can graph it like the following https://www.desmos.com/calculator/mxjrmz7abx Ok, well that is my take on the subject. A lot to wrap your brain around :-) . Good luck

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