Asked by Natalia
I am supposed to write the standard equation of the parabola with the directrix x=1, and the vertex 6,2. I got y-6=1/4p(x-2)^2. Is this correct?
I am supposed to graph y+3=-1/12(x-1)^2. I tried to find p, and I got 36. I don't think this could be correct, because the graph I am given does not have that high of a range. THe graph almost forms your traditional cross, with just a little bit of space in the positive quadrants, and a lot of space in the negative quadrants. What is the real definition of p? How do you find that? Then what is the focus and the directrix from p? Thanks
I am supposed to graph y+3=-1/12(x-1)^2. I tried to find p, and I got 36. I don't think this could be correct, because the graph I am given does not have that high of a range. THe graph almost forms your traditional cross, with just a little bit of space in the positive quadrants, and a lot of space in the negative quadrants. What is the real definition of p? How do you find that? Then what is the focus and the directrix from p? Thanks
Answers
Answered by
North
I'll answer the first question.
The directrix is vertical so the parabola opens sideways. The equation is therefore of the form:
4p(x - h) = (y - k)^2
The directed distance from the directrix to the vertex is p.
p = 6 - 1 = 5
4p = 4*5 = 20
p is positive so the parabola opens to the right.
The equation of the parabola is therefore:
4p(x - h) = (y - k)^2
20(x - 6) = (y - 2)^2
The directrix is vertical so the parabola opens sideways. The equation is therefore of the form:
4p(x - h) = (y - k)^2
The directed distance from the directrix to the vertex is p.
p = 6 - 1 = 5
4p = 4*5 = 20
p is positive so the parabola opens to the right.
The equation of the parabola is therefore:
4p(x - h) = (y - k)^2
20(x - 6) = (y - 2)^2