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K.Binks
 one year ago
How do you graph the focus (0,8) of a parabola?
K.Binks
 one year ago
How do you graph the focus (0,8) of a parabola?

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dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0What do you mean when you say "focus"? Are you talking about the vertex, or the yintercept?

K.Binks
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

K.Binks
 one year ago
Best ResponseYou've already chosen the best response.0they didnt explain, they just told me what it was (0,8)

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I do not know how to do that. Maybe @triciaal can help?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0another reason not to accept answers without understanding I was thinking "doesn't this person know that the focus is just a point"

K.Binks
 one year ago
Best ResponseYou've already chosen the best response.0No, I didn't know that, which is why I was asking someone to explain

dessyj1
 one year ago
Best ResponseYou've already chosen the best response.0Sorry about that. She was talking to me.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438625611026:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0the focus to the vertex is the same distance of the vertex to the directrix

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@K.Binks do you understand?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0y = 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
 one year ago
Best ResponseYou've already chosen the best response.0don't want to overload you are you understanding this so far?

K.Binks
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I get that so far

K.Binks
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0Sorry 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
 one year ago
Best ResponseYou've already chosen the best response.0The directrix = y = 1/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyways, 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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