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

1. anonymous

Hi.

2. anonymous

What do you mean when you say "focus"? Are you talking about the vertex, or the y-intercept?

3. 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.

4. K.Binks

they didnt explain, they just told me what it was (0,8)

5. anonymous

Yeah, I do not know how to do that. Maybe @triciaal can help?

6. triciaal

another reason not to accept answers without understanding I was thinking "doesn't this person know that the focus is just a point"

7. K.Binks

No, I didn't know that, which is why I was asking someone to explain

8. anonymous

Sorry about that. She was talking to me.

9. 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

10. triciaal

|dw:1438625611026:dw|

11. triciaal

the focus to the vertex is the same distance of the vertex to the directrix

12. triciaal

13. anonymous

@K.Binks do you understand?

14. 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

15. triciaal

don't want to overload you are you understanding this so far?

16. K.Binks

Yes, I get that so far

17. 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

18. 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.

19. anonymous
20. anonymous

The directrix = y = -1/4

21. 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